L-functions of symmetric powers of cubic exponential sums

L-functions of symmetric powers of cubic exponential sums C. Douglas Haessig February 2, 2008

arXiv:math/0608521v4 [math.NT] 9 Jan 2008

Abstract For each positive integer k, we investigate the L-function attached to the k-th symmetric power of the F -crystal associated to the family of cubic exponential sums of x3 + λx where λ runs over Fp . We explore its rationality, field of definition, degree, trivial factors, functional equation, and Newton polygon. The paper is essentially self-contained, due to the remarkable and attractive nature of Dwork’s p-adic theory. A novel feature of this paper is an extension of Dwork’s effective decomposition theory when k < p. This allows for explicit computations in the associated p-adic cohomology. In particular, the action of Frobenius on the (primitive) cohomology spaces may be explicitly studied.

1991 Mathematics Subject Classification: Primary 11L99, 14F30

Contents 1 Introduction

1

2 Relative Dwork Theory

4

3 Fibres

12

4 Variation of Cohomology (Deformation Theory)

17

5 Cohomological Formula for Mk (T )

22

6 Cohomology of Symmetric Powers

25

7 Dual Cohomology of Symmetric Powers

32

8 Newton Polygon of Mk (T )

36

1

Introduction

This paper represents a continuation in the study of L-functions attached to symmetric powers of families of exponential sums. Our central object of study will be the family of cubic exponential sums of x3 + λx. Similar studies have considered the Legendre family of elliptic curves [1] [8], and the n-variable Kloosterman family [10] [13]. Various approaches have been used to study these L-functions. Dwork [8] used the existence of a Tate-Deligne mapping (excellent lifting) of the elliptic family to study the symmetric powers; this line of investigation was continued by Adolphson [1] in his thesis. Another related method is the symmetric power of the associated F -crystal. This approach was explored by Robba [13] who used index theory to calculate the degrees of such functions coming from the family of one-variable Kloosterman sums. A third approach, developed by Fu and Wan [10], used ℓ-adic methods of Deligne and Katz to study Kloosterman sums in n-variables. Since we are interested in the p-adic properties of the zeros and poles, and we believe excellent lifting does not exist for our family when p ≡ −1 mod(3), we take an approach similar to that of Robba’s. The L-function, denoted Mk (T ), attached to the k-th symmetric power of the cubic family is defined as follows. Fix a prime p ≥ 5 and let ζp ∈ Cp be a primitive p-th root of unity. Let λ be an element of Fp and denote by deg(λ) := [Fp (λ) : Fp ] its degree. Define q := pdeg(λ) . The L-function associated to the cubic exponential sums Sm (λ) :=

X

T rFqm /Fp (x3 +λx)

ζp

x∈Fqm

1

m = 1, 2, 3, . . .

is well-known to be a quadratic polynomial with coefficients in Z[ζp ]:   m X T  = (1 − π1 (λ)T )(1 − π2 (λ)T ). L(λ, T ) := exp  Sm (λ) m m≥1

The L-function of the k-th symmetric power of this family is defined by: k Y Y

Mk (T ) :=

λ∈|A1 | i=0

(1 − π1 (λ)i π2 (λ)k−i T deg(λ) )−1

where |A1 | denotes the set of Zariski closed points of A1 . The main theorem of this paper is the following. Theorem 1.1. Let k be a positive integer. 1. Mk (T ) is a rational function with coefficients in Z[ζp ]. Further, if p ≡ 1 mod(3), then the coefficients of Mk (T ) lie in the ring of integers of the unique subfield L of Q(ζp ) with [L : Q] = 3. However, if p ≡ −1 mod(3) then Mk (T ) has integer coefficients. 2. For k odd, Mk (T ) is a polynomial and satisfies the functional equation   Mk (T ) = cT δ Mk p−(k+1) T −1

(1.1)

where c is a nonzero constant and δ := deg Mk . Furthermore, if k is odd and k < p, then Mk (T ) has degree (k + 1)/2, and writing Mk (T ) = 1 + c1 T + c2 T + · · · + c(k+1)/2 T (k+1)/2 ,

a quadratic lower bound for the Newton polygon is given by: ordp (cm ) ≥

(p − 1)2 2 (m + (k + 1)m) 3p2

for m = 0, 1, . . . , (k + 1)/2. 3. For k even, we may factorize Mk (T ) as follows: ( fk (T ) Nk (T )M Mk (T ) = Nk (T )M fk (T )

for k even and k < 2p for k even and k ≥ 2p

Qk (T )

fk (T ) has degree ≤ k and satisfies the functional equation (1.1), and where M Nk (T ) = (1 − pk/2 T )mk k/2

= (1 − (−¯ g)

= (1 − p

k/2

if p ≡ 1 mod(12)

mk

T)

nk

T ) (1 + p

k/2

mk −nk

T)

= (1 − pk/2 T )mk −nk (1 + pk/2 T )nk k/2

T ) (1 + g¯

k/2

mk −nk

= (1 − g¯

= (1 − g¯

where

nk

T)

 j k 1 + k 2p mk := j k  k 2p

k/2

mk −nk

T)

k/2

(1 + g¯

if 4|k

T)

and

if 4 ∤ k

nk

if p ≡ 5 mod(12)

if 4|k and p ≡ 7 mod(12)

if 4 ∤ k and p ≡ 7 mod(12) if 4|k and p ≡ 11 mod(12)

if 4 ∤ k and p ≡ 11 mod(12)

 j k 1 + k 4p nk := j k  k 4p

if 4|k if 4 ∤ k

and where g¯ is the complex conjugate of the Gauss sum g := g2 ((p2 − 1)/3) (see §4). Note, we conjecture Qk (T ) always equals 1; see the conjecture below. The lower bound for the Newton polygon is obtained using an extension of Dwork’s “General Theory” (see [5]). This theory produces an explicit algorithm to compute elements in cohomology. More specifically, given an (k) analytic element ξ on the level of the cochain (Ma (b′ , b) in our notation) with known growth rate, Theorem 6.3 ¯ describes the growth rate for the reduction ξ of this element in the cohomolgy Hk1 . Since we are able to easily 2

understand the Frobenius action on the cochain level in terms of growth rates, we can translate this into growth rates in cohomology. The lower bound of the Newton polygon follows (see Theorem 8.1). We expect the lower bound may be improved to 13 (m2 + m + mk). This would be optimal since, for k = 3 and p = 7, Vasily Golyshev has computed ordp (c1 ) = 5/3 which coincides with this predicted lower bound. This expected lower bound follows from the general philosophy that as p → ∞, the Newton polygon equals the Hodge polygon (see [17, Conjecture 1.9] [20]). (Note, the k3 m term is just an artifact leftover from the divisibility of the Frobenius matrix by pk/3 .) We believe this lower bound will also hold for k even and k < p. Looking for lower bounds of this type are inspired by a reciprocity theorem of Wan’s [16]. The Govˆea-Mazur conjecture predicts a certain (essentially linear) upper bound related to the arithmetic variation of dimensions associated to classical modular forms of weight k on Γ1 (N ) ∩ Γ0 (p), with (N, p) = 1. To prove a quadratic upper bound for this conjecture, Wan [16] proves a reciprocity theorem which cleverly transforms a quadratic uniform lower bound for the Newton polygon associated to an Atkin’s operator on a space of p-adic modular forms into a quadratic upper bound for the Govˆea-Mazur conjecture. At this time, we are only able to prove a lower bound for the Newton polygon when k is odd and k < p. The reason for the second restriction comes from certain rational numbers in the proof of our decomposition theorem (Theorem 6.3). Their denominators are p-adic units when k < p, but are often not units when k ≥ p. We believe the obstruction to proving a similar bound for k even and k < p lies in the underlying variety, or motive, associated to the k-th symmetric power, being singular. Evidence for this is given in Livn´e [11]. In that paper, Livn´e relates the k-th moment of the cubic family ax3 + bx to Fp -rational points on a hypersurface Wk in Pk−2 . When k is odd, Wk is nonsingular, but when k is even, Wk has ordinary double points. Livn´e overcomes this difficulty by embedding Wk in a non-singular family. Interestingly, Dwork proceeds in a similar manner when trying to generalize his effective decomposition theory [5, §3d], which works well for non-singular projective hypersurfaces, to singular ones [6]. We are curious whether one can adapt these methods to the current situation. A first step in this direction may lie in a partial effective decomposition theorem (Theorem 6.12) proven when k is even and k < p. Even though the theorem does not give complete information about the cohomology, it does produce several non-trivial basis vectors of the cohomology and some information concerning the lower bound of the Newton polygon. Another topic is the denominator of Mk (T ). As we shall see in §5 and §7.1, Mk (T ) is a quotient of characteristic polynomials of Frobenius matrices acting on cohomology spaces: Mk (T ) =

Nk (T )det(1 − β¯k T |P Hk1 ) . det(1 − pβ¯k T |Hk0 )

(1.2)

The polynomial Qk (T ) is defined as det(1 − pβ¯k T |Hk0 ). For every odd k, or for every even k with k < 2p, Hk0 has dimension zero. We suspect this is always true: Conjecture 1.2. Hk0 = 0 for all positive integers k. Consequently, Mk (T ) is a polynomial for all positive integers k. Our methods also allow us to study the L-function, denoted Mk (d, T ), attached to the symmetric powers of the family xd + λx, when d is not divisible by p. In particular, we are able to prove: Theorem 1.3. Mk (d, T ) is a rational function with coefficients in Z[ζp ]. If r := (d, p − 1) then the coefficients of Mk (d, T ) lie in the ring of integers of the unique subfield L of Q(ζp ) with [L : Q] = r. In particular, if (d, p−1) = 1 then Mk (d, T ) has integer coefficients. We begin the paper by defining a relative cohomology theory (the so-called d-Airy F -crystal) tailored specifically for the family xd + ax, where a is the parameter. This will be a free module Ma (b′ , b) of rank d over a power series ring L(b′ ). It will carry an action of Frobenius α(a) and a connection ∂a . As we shall see in §3.1, when we ∗ specialize the parameter a to a lifting in Qp of an element z¯ ∈ Fp , the relative cohomology reduces to a Cp -vector space Mz , and the L-function of xd + z¯x has the cohomological description as the characteristic polynomial of the Frobenius on Mz : L(xd + z¯x, T ) = det(1 − α ¯ z,s T |Mz ) We then prove explicitly the functional equation for this L-function via Dwork’s theory. This means defining a relative dual space R′−π,a (b′ , b) to Mπ,a (b′ , b), and an isomorphism Θ−π,a between them which relates the Frobenius α ¯π (a) to its conjugate dual operator α ¯ ∗−π (a): ¯ π (a) ◦ Θ−π,a ◦ α Θ−π,ap = p−1 α ¯∗−π (a). Lastly, we find the exact p-adic order of the entries in the Frobenius matrix α ¯ (a). This allows us to explicitly describe the Newton polygon of L(xd + z¯x, T ); see §3.1. 3

Next, for d = 3, the connection ∂a on the relative cohomology produces a system of differential equations, the Airy differential system, which has only an irregular singular point at infinity. The (dual) Frobenius is an isomorphism on the local solutions of this system and so, locally, is given by an invertible constant matrix M . For local solutions near 0, M may be explicitly described in terms of Gauss sums. When the solutions are near infinity, M may be explicitly described in terms of the square roots of p or Gauss sums g; see §4.1 for details. The description of M depends on the congruence class of p modulo 12 and is the main reason for so many different forms of Nk (T ) in Theorem 1.1. In §5, we present the general theory for a cohomological formula for Mk (T ) following the work of Robba. In (k) short, the k-th symmetric power of the free module Ma (b′ , b) is another free module Ma (b′ , b) which carries a new action of Frobenius βk which is built from the k-th symmetric power of the Frobenius α(a), ¯ and a new connection which is also denoted ∂a . This means we are able to take cohomology once again, creating the finite dimensional Cp -vector spaces Hk0 and Hk1 . The cohomological formula for Mk (T ) is equation (1.2) above. In §7.1, we shall see that the dual space of Hk1 splits into three parts: a constant subspace Ck+1 , a trivial p subspace Tk , and a primitive part P Hk1∗ . For the first two, the action of Frobenius is explicitly described. These descriptions lead to the polynomials Pk (T ) and Nk (T ) described in Theorem 1.1. The action of the Frobenius on P Hk1∗ is more difficult to understand, however, we are able to present the theory for the functional equation of fk (T ). This is similar in nature to that of L(xd + z¯x, T ), yet different since the analogous operator to that of M Θ−π,a has a kernel which must be dealt with. In §6, we present an effective decomposition theory for the cohomology space Hk1 when k is odd and k < p. (k) More precisely, an explicit procedure is described which takes an element ξ of Ma (b) and produces its reduction (k) (k) ξ¯ in Hk1 = Ma (b)/∂a Ma (b). As an application, we may compute a non-trivial lower bound for the entries of ¯ the Frobenius matrix βk acting on P Hk1 . This produces the quadratic lower bound for the Newton polygon given in Theorem 1.1; see §8. Acknowledgements. I wish to thank Alan Adolphson for many helpful comments, Steve Sperber for useful conversations, Zhi-Wei Sun for his proof of the combinatorial formula (6.4), Vasily Golyshev for pointing out some technical errors, and Daqing Wan for many insightful conversations. Lastly, I would like to thank the referees for suggesting many improvements and corrections.

2 2.1

Relative Dwork Theory Relative Cohomology

In this section we define and study a cohomology theory specifically suited for the more general family xd + ax with p ∤ d. While the growth conditions on the Banach spaces L(b′ ; ρ) and K(b′ , b; ρ) below may seem elaborate, they will allow us to obtain a detailed description of the associated cohomology, as well as, provide us with an efficient means of reduction modulo the operator Da . As an arithmetic application, in §2.6 we are able to explicitly study the action of Frobenius on the relative cohomology space Ha (b′ , b) defined below. Let d be a positive integer relatively prime to p. Let b and b′ be two positive real numbers. We will assume throughout this section that b ≥ b′ > 0. With ρ ∈ R, define the spaces L(b′ ; ρ) := { ′

L(b ) :=

∞ X i=0

[

Bi ai : ord(Bi ) ≥ b′ (1 − 1/d)i + ρ, ∀i ≥ 0}

L(b′ ; ρ)

ρ∈R ′

K(b , b; ρ) := { ′

K(b , b) := ′

V(b , b; ρ) := ′

V(b , b) :=

X

i,j≥0

[

ρ∈R

Bij ai xj : ord(Bij ) ≥ b′ (1 − 1/d)i + bj/d + ρ, ∀i ≥ 0}

K(b′ , b; ρ)

d−1 M

′

L(b ; ρ)x

i=0

[

ρ∈R

i

!

∩ K(b′ , b; ρ)

V(b′ , b; ρ).

Notice that K(b′ , b) is an L(b′ )-module.

4

Fix π ∈ Cp such that π p−1 = −p. Define Da , an L(b′ )-module endomorphism of K(b′ , b), by Da := x

∂ + π(dxd + ax). ∂x

It is useful to keep in mind that, formally, Da = e−π(x

d

+ax)

◦x

d ∂ ◦ eπ(x +ax) . ∂x

(2.1)

d

Note, we need to use the word “formally” because multiplication by eπ(x +ax) is not an endomorphism of K(b′ , b). Using this, we may define a relative cohomology space as the L(b′ )-module Ha (b′ , b) := K(b′ , b)/Da K(b′ , b). As the following theorem demonstrates, Ha (b′ , b) is a free L(b′ )-module of rank d. Theorem 2.1. Let b be a positive real number such that e := b −

1 p−1

> 0. Then, for every ρ ∈ R we have

K(b′ , b; ρ) = V(b′ , b; ρ) ⊕ Da K(b′ , b; ρ + e) and K(b′ , b) = V(b′ , b) ⊕ Da K(b′ , b)

Furthermore, Da is an injective operator on K(b′ , b).

Remark 2.2. We will prove this theorem using Dwork’s general method [5] which consists of the following six lemmas. Note that the last three lemmas follow automatically from the validity of the first three. Lemma 2.3. Define H := π(dxd + ax). Then for every ρ ∈ R K(b′ , b; ρ) = V(b′ , b; ρ) + HK(b′ , b; ρ + e). P Proof. Note, it is sufficient to prove this for ρ = 0. Let u = n,s≥0 Bs,n as xn ∈ K(b′ , b; 0). We may write this as u=

X

Cn xn

where Cn :=

n≥0

For each j ≥ 0, define

X

Bs,n as .

s≥0

Aj :=

X

(−1)i

i≥0

 a i d

Cdi+d+(j−i) .

With A−1 := 0, a calculation shows that u = P + HQ where P :=

d−1 X a (Cj − Aj−1 )xj d j=0

and Q :=

1 X Aj xj . dπ j≥0

It is not hard to show P (a, x) ∈ K(b′ , b; 0) and Q(a, x) ∈ K(b′ , b; e). Lemma 2.4. HK(b′ , b) ∩ V(b′ , b) = {0}. Proof. Suppose π(dxd + ax)

X j≥0

P

j

′

Bj xj = C0 + C1 x + · · · + Cd−1 xd−1

′

where j≥0 Bj x ∈ K(b , b), Bj ∈ L(b ), and the right-hand side is in V(b′ , b). For each j ≥ 0, the coefficient of xd+j on the left-hand side of the above equals zero: πdBj + πaBd+j−1 = 0 Iterating this i times yields Bj =



=⇒

−a d

i

a Bj = − Bd+j−1 . d

Bid−i+j .

Thus, Bj is divisible by ai for all i ≥ 1, and so Bj must equal zero. 5

Lemma 2.5. If f ∈ K(b′ , b) and Hf ∈ K(b′ , b; ρ), then f ∈ K(b′ , b; ρ + e). P P Proof. Write f = j≥0 Aj xj where Aj := i≥0 Gij ai . Now, X

H

Aj xj =

j≥0

where Cj :=

P

i≥0

X j≥0

Cj xj ∈ K(b′ , b; ρ)

Bij ai . In particular, this means X

[dAj−d + aAj−1 ]xj =

1X Cj xj . π j≥d

j≥d

Thus,

a 1 Aj = − Aj+d−1 + Cj+d for every j ≥ 0. d dπ Iterating the A’s on the right-hand side N times, we obtain the formula: Aj =

N −1 X (−a)i 1 (−a)N A + Cj+d(i+1)−i . j+N (d−1) N d di dπ i=0

Specializing a with ord(a) + b′ (1 − d1 ) > 0, then finding δ ∈ R such that f ∈ K(b′ , b; δ) we have ord(

(−a)N 1 Aj+N (d−1) ) ≥ N [ord(a) + b(1 − )] + bj/d + δ. dN d

Since b ≥ b′ , the coefficient of N is positive. Thus, for a specialized, letting N tend to infinity we have: X  −a i 1 Cj+d(i+1)−i d dπ i≥0 X = Gr,j ar

Aj =

r≥0

where Gr,j := It follows that f (a, x) =

P

r X i=0

j,r≥0

r j

(−1)i

1 Br−i,j+d(i+1)−i . πdi+1

Gr,j a x ∈ K(b′ , b; ρ + e).

Lemma 2.6. For every ρ ∈ R, K(b′ , b; ρ) = V(b′ , b; ρ) + Da K(b′ , b; ρ + e). Proof. It is sufficient to prove this for ρ = 0. Let f ∈ K(b′ , b; 0). Set f (0) := f . Then there exists a unique η (0) ∈ V(b′ , b; 0) and ξ (0) ∈ K(b′ , b; e) such that f (0) = η (0) + Hξ (0) . Define f (1) : = f (0) − η (0) − Da ξ (0) ∂ = −x ξ (0) ∈ K(b′ , b; e). ∂x Then, there exists a unique η (1) ∈ V(b′ , b; e) and ξ (1) ∈ K(b′ , b; 2e) such that f (1) = η (1) + Hξ (1) . Define f (2) := f (1) − η (1) − Da ξ (1) . Continuing h times we get: f (h) := f (h−1) − η (h−1) − Da ξ (h−1) .

6

Adding all these together, we obtain f (h) = f (0) −

h−1 X i=0

η (i) − Da

h−1 X i=0

ξ (i) ∈ K(b′ , b; he).

Thus, as h → ∞, f (h) → 0 in H(b′ , b), leaving X X f= η (i) + Da ξ (i) ∈ V(b′ , b) + Da K(b′ , b). i≥0

i≥0

Lemma 2.7. If f ∈ K(b′ , b) and Da f ∈ K(b′ , b; ρ), then f ∈ K(b′ , b; ρ + e). Proof. If f 6= 0, then we may choose c ∈ R such that f ∈ K(b′ , b; c) but f ∈ / K(b′ , b; c + e). By hypothesis, ′ Da f ∈ K(b , b; ρ), thus ∂ Hf = Da f − x f ∈ K(b′ , b; ρ) + K(b′ , b; c) = K(b′ , b; l) ∂x where l := min{ρ, c}. Thus, f ∈ K(b′ , b; l + e) which means l 6= c. Hence l = ρ as desired. Corollary 2.8. ker(Da |K(b′ , b)) = 0. Proof. Suppose Da (f ) = 0. Since 0 ∈ K(b′ , b; ρ) for all ρ ∈ R, by the previous lemma, f ∈ K(b′ , b; ρ + e) for all ρ ∈ R. However, the only element with this property is 0. Lemma 2.9. Da K(b′ , b) ∩ V(b′ , b) = {0}. Proof. Let f ∈ Da K(b′ , b) ∩ V(b′ , b) and suppose f 6= 0. Choose ρ ∈ R such that f ∈ K(b′ , b; ρ) but f ∈ / K(b′ , b; ρ + e). Also, let η ∈ K(b′ , b) such that Da η = f . By Lemma 2.7, η ∈ K(b′ , b; ρ + e) and so f − Hη = ∂ η ∈ K(b′ , b; ρ + e). Thus, there exists ζ ∈ V(b′ , b; ρ + e) and w ∈ K(b′ , b; ρ + 2e) such that x ∂x x

∂ η = ζ + Hw. ∂x

This means f = ζ + H(η + w). Since f ∈ V(b′ , b) we must have f = ζ ∈ V(b′ , b; ρ + e). But this contradicts our choice of ρ. Remark 2.10. Notice that in the proof of Lemma 2.3 that if u is divisible by x, then P is also divisible by x but QPneed not be. P From this and the proof of Lemma 2.6, if f ∈ K(b′ , b; ρ) is divisible by x then when P we write (i) η (i) is f = ≥0 η + Da i≥0 ξ (i) in the decomposition of K(b′ , b; ρ) = V(b′ , b; ρ) ⊕ Da K(b′ , b; ρ + e), then P (i) divisible by x but ξ need not be. This will be important when we define the relative primitive cohomology in §2.4.

2.2

Relative Dual Cohomology

′ ′ ′ ′ In the last section, we saw that {xi }d−1 i=0 is a basis of the L(b )-module Ha (b , b) := K(b , b)/Da K(b , b). In this ′ section, we will describe a dual space of Ha (b , b) and compute its dual basis via a nondegenerate pairing. Let b∗ be a positive real number with b > b∗ . Define the L(b′ )-module   ∞  X 1 Bij ai x−j | inf (ord(Bij ) − (b′ (1 − )i − b∗ j/d)) > −∞ . K∗ (b′ , b∗ ) := i,j   d i,j=0

Define the pairing h·, ·i : K(b′ , b) × K∗ (b′ , b∗ ) → L(b′ ) as follows. With g ∗ ∈ K∗ (b′ , b∗ ) and h ∈ K(b′ , b), thinking of the variable a as a constant, define

hh, g ∗ i := constant term w.r.t. x in the product hg ∗ . P −j ∈ H∗ (b′ , b∗ ) and h(a, x) := For clarity, we will write this out explicitly: with g ∗ (a, x) := j≥0 Aj (a)x P j ′ j≥0 Bj (a)x ∈ K(b , b), ∞ ∞ X X X hh, g ∗ i := Gs as where Gs := Air Bjr . s=0

r=0 i+j=s

7

This defines a perfect pairing (or nondegenerate pairing) in the sense that no nonzero element g ∗ ∈ K∗ (b′ , b∗ ) exists such that hh, g ∗ i = 0 for every h ∈ K(b′ , b) and vice versa. Define the truncation operator T runcx : Cp [[a, x±1 ]] → Cp [[a, x−1 ]] by linearly extending ( xn if n ≤ 0 n T runcx (x ) := . 0 if n > 0 Define the K∗ (b′ , b∗ )-endomorphism

∂ + T runcx [π(dxd + ax)] ∂x d d ∂ ◦ e−π(x +ax) ]. (2.2) = T runcx [eπ(x +ax) ◦ −x ∂x The operators Da and Da∗ are dual to one another with respect of the pairing. That is, for every g ∗ ∈ K∗ (b′ , b∗ ) and h ∈ K(b′ , b), we have hh, Da∗ g ∗ i = hDa h, g ∗ i. Da∗ : = −x

To see this, notice that the operator T runcx does not affect constant terms (with respect to x), and so, ∂ ∂ T runcx [π(dxd + ax)] is dual to π(dxd + ax). Next, that −x ∂x is dual to x ∂x follows from the Leibniz identity d d d x (hg ∗ ) = x (h)g ∗ + hx (g ∗ ) dx dx dx since the left-hand side has no constant term. Define the L(b′ )-module Ra (b′ , b∗ ) := ker(Da∗ |K∗ (b′ , b∗ )). We wish to show that Ra (b′ , b∗ ) is the dual space of Ha (b′ , b).

1 Theorem 2.11. Let b > p−1 > b′ > 0, and b ≥ b∗ > 0. Then Ra (b′ , b∗ ) is the algebraic dual of Ha (b′ , b). d−1 ′ ∗ ′ ∗ Furthermore, dual to the basis {xi }d−1 i=0 of Ha (b , b) is the basis {gi (a)}i=0 ⊂ Ra (b , b ) which takes the explicit ∗ form: g0 = 1, and for i = 1, 2, . . . , d − 1

gi∗ (a)

=

i+d−2 ∞ ⌊ X X r=i

l=0

l−(r−i) d

X ⌋ j=0

Z (dπ)(d−1)j+(r−i)



−a d

l−(r−i)−dj

x−((d−1)l+r) .

where “Z” indicates some determinable nonzero integer. Proof. It is clear that g0∗ = 1 is the dual vector of 1 ∈ Ha (b′ , b). For i = 1, . . . , d − 1, let gi∗ (a) := x−i + P (i) (i) −j j≥d Bj x . Let us determine these Bj . (i)

From Da∗ (gi∗ (a)) = 0, the coefficients Bj

must satisfy the recurrence relation

(i)

Bd+j =

−j (i) a (i) B − Bj+1 . dπ j d

Thus, from the initial conditions given by requiring hxj , gi∗ (a)i = δij , where δij is the Kronecker delta symbol, (i) each Bj may be uniquely solved by a series of the form indicated above. Next, notice that the p-adic order of the coefficient of al−(r−i)−dj x−((d−1)l+r) in gi∗ (a) is bounded below by (d−1)j+(r−i) 1 since d is a p-adic unit. Since b ≥ p−1 ≥ b′ > 0, we have − p−1   b 1 (d − 1)j + (r − i) ′ (l − (r − i) − dj) − ((d − 1)l + r). − ≥b 1− p−1 d d

This means gi∗ (a) ∈ K∗ (b′ , b∗ ; 0). ′ ∗ ′ ∗ To show {gi∗ (a)}d−1 i=0 is a basis of Ra (b , b ), we need only show that they span Ra (b , b ) since it is clear that ∗ ′ ∗ they are linearly independent. This is demonstrated as follows: Let h ∈ Ra (b , b ) and define hi := hxi , h∗ i for P −j ∗ ˜∗ ˜∗ ˜∗ P , each i = 0, 1, . . . , d−1. Next, define ˜ h∗ := h∗ − d−1 j≥0 hj (a)x i=0 hi gi (a). We wish to show h = 0. Now, h = n+d n n n+1 ∗ ˜ = Da (x ) − nx − πax and by construction, hi = 0 for i = 0, 1, . . . , d − 1. Using the reduction formula dπx ˜ ∗ ) = 0, it is easy to see that and Da∗ (h ˜ ∗ = hxd+n , h ˜ ∗i = h n+d ′

˜∗

where Ai ∈ L(b ). Hence, h = 0 as desired. 8

d−1 X i=0

Ai ˜h∗i = 0

2.3

Relative Dwork Operators

In this section, we define the Dwork operator α(a) ¯ on the relative cohomology Ha (b′ , b), and its dual α ¯ ∗ (a) on p−1 1 b ′ ∗ ′ Ra (b , b ). Throughout this section we will fix real numbers b and b such that p ≥ b > p−1 and p ≥ b′ > 0. Fix π ∈ Cp such that π p−1 = −p. P∞ p Dwork’s first splitting function on Fp is θ(t) := eπ(t−t ) = i=0 θi ti . It is well-known [12] that ord(θi ) ≥ p−1 p2 i πi i!

for every i. Also, θi =

for each i = 0, 1, 2, . . . , p − 1. Next, define F (a, x) :=

Writing F (a, x) =

exp π(xd + ax) = θ(xd )θ(ax). exp π(xdp + ap xp )

P⌊r/d⌋ r r−di , then the r≥0 Hr (a)x with Hr (a) := i=0 θi θr−di a (b/p) 1 ′ d )(r − di) + d r. Consequently, F (a, x) ∈ K(b , b/p). It ′

P

ord(θi θr−di ) ≥ b′ (1 − F (a, x) is an endomorphism of K(b , b/p). From the Cartier operator defined as

ψx : K(b′ , b/p) → K(b′ , b) takes

∞ X i=0

Bi (a)xi 7−→

∞ X

coefficient of ar−di xr satisfies follows that multiplication by

Bpi (a)xi ,

i=0

we may define the Dwork operator α(a) := ψx ◦ F (a, x) : K(b′ , b) → K(b′ , b). It is useful to keep in mind the formal identity: α(a) = e−π(x

d

+ap x)

◦ ψx ◦ eπ(x

d

+ax)

.

(2.3)

∂ ∂ Since px ∂x ◦ ψx = ψx ◦ x ∂x , and for any u ∈ Cp [[x]], u(x) ◦ ψx = ψx ◦ u(xp ), equations (2.1) and (2.3) demonstrate that α(a) ◦ Da = pDap ◦ α(a). Consequently, α(a) induces two L(b′ )-linear maps on the relative ¯ : ker(Da |K(b′ , b)) → ker(Dap |K(b′ , b)). cohomology: α ¯ (a) : Ha (b′ , b) → Hap (b′ , b) and α(a)

Remark 2.12. Let us show α ¯ (a) is an isomorphism by creating a right-inverse on the co-chain level which will reduce to an isomorphism on cohomology. Denote by F−π (a, x) the replacement of π by its conjugate −π in F (a, x). Notice that F−π (a, x) = F (a, x)−1 ∈ K(b′ , b/p). Define the Frobenius map Φx : Cp [[a, x±1 ]] → Cp [[a, x±p ]] by linearly extending

x 7→ xp .

Next, define the endomorphism α(a)−1 := F−π (a, x) ◦ Φx : K(b′ , b) → K(b′ , b) and note α(a)−1 = e−π(x

d

+ax)

◦ Φx ◦ eπ(x

d

+ap x)

.

From this and (2.3), α(a) ◦ α(a)−1 = id. Warning, α(a)−1 is not a left inverse of α(a). ∂ ∂ = x ∂x ◦ Φx , we see that Next, from (2.1) and the relation pΦx ◦ x ∂x pα(a)−1 ◦ Dap = Da ◦ α(a)−1 . ¯ (a)¯ α(a)−1 = id. Since these Hence, α(a)−1 induces a mapping α(a) ¯ −1 : Hap (b′ , b) → Ha (b′ , b) which satisfies α ′ maps are acting on free L(b )-modules of finite rank, α(a) ¯ must be an isomorphism. Dual Dwork Operator. Define the operator α∗ (a) : K∗ (b′ , b∗ ) → K∗ (b′ , b∗ ) by the following: α∗ (a) : = T runcx [F (a, x) ◦ Φx ] = T runcx [eπ(x

d

+ax)

◦ Φx ◦ e−π(x

d

+ap x)

].

(2.4)

Notice that ψx and Φx are dual operators in the sense that for every g ∗ ∈ K∗ (b′ , b∗ ) and h ∈ K(b′ , b), we have hψx h, g ∗ i = hh, Φx g ∗ i. Next, since the operator T runcx does not kill any constant terms (with respect to x), T runcx [F (a, x)] is dual to F (a, x). Putting these together yields the duality between α∗ (a) and α(a). By taking the dual of α(a) ◦ Da = pDap ◦ α(a), we have pα∗ (a) ◦ Da∗p = Da∗ ◦ α∗ (a). Therefore, α∗ (a) induces a mapping ¯ (a) : Ha (b′ , b) → Hap (b′ , b) and α∗ (a) : Rap (b′ , b∗ ) → Ra (b′ , b∗ ) are α∗ (a) : Rap (b′ , b∗ ) → Ra (b′ , b∗ ). Moreover, α dual to one another with respect to the pairing h·, ·i. 9

2.4

Relative Primitive Cohomology and its Dual

Define the L(b′ )-module

xK(b′ , b) := {g ∈ K(b′ , b)|g is divisible by x}.

Notice that α(a) and Da are well-defined endomorphism of xK(b′ , b). Define Ma (b′ , b) := xK(b′ , b)/Da K(b′ , b). From Theorem 2.1 and Remark 2.10, Ma (b′ , b) is a free L(b′ )-module with basis {xi }d−1 i=1 . Since we still have pDap ◦ α(a) = α(a) ◦ Da , α(a) induces a mapping α ¯ (a) : Ma (b′ , b) → Map (b′ , b). Using a pairing identical to that between Ha (b′ , b) and Ra (b′ , b∗ ), we see that the subspace generated by 1 in Ra (b′ , b∗ ) is the annihilator of Ma (b′ , b). Hence, the dual of Ma (b′ , b) is the free L(b′ )-module R′a (b′ , b∗ ) := Ra (b′ , b∗ )/h1i with basis {gi∗ (a)}d−1 ¯ (a) is α ¯ ∗ (a), induced by α∗ (a), and this map takes R′ap (b′ , b∗ ) i=1 . Dual to α ′ ′ ∗ ∗ bijectively onto Ra (b , b ). Notice that α ¯ (a)(1) = 1.

2.5

Relative Functional Equation

So far, we have fixed a solution π of the equation xp−1 = −p. Since we are assuming p is odd, −π is also a solution. Let us see how we would have proceeded had we used −π. But before we do this, we need to modify our current notation to keep things clear. Let us denote the operator Da by Dπ,a , the space Ma (b′ , b) by Mπ,a (b′ , b), and the Dwork operator α ¯ (a) by α ¯ π (a). Also, denote by Rπ,a (b′ , b∗ ) and α∗π (a) the dual space Ra (b′ , b∗ ) and dual Dwork operator α∗ (a). ∂ − π(dxd + ax), then H−π,a (b′ , b) := K(b′ , b)/D−π,a K(b′ , b) would still be a Now, had we used D−π,a := x ∂x ′ i d−1 free L(b )-module with basis {x }i=0 . Let M−π,a (b′ , b) := xK(b′ , b)/D−π,aK(b′ , b). Using the same pairing as that between Hπ,a (b′ , b) and Rπ (b′ , b∗ ), the dual of H−π,a (b′ , b) is the space ∗ ∗ R−π,a (b′ , b∗ ) := ker(D−π,a |K∗ (b′ , b∗ )). Clearly, a basis for the latter space is {g−π,i }d−1 i=0 , where we have replaced π for −π in the formulas given in Theorem 2.11. Next, dual to α−π (a) := ψx ◦ F−π (a, x) is α∗−π (a) := T runcx ◦ F−π (a, x) ◦ Φx . Notice that α∗−π (a) : R−π,ap (b′ , b∗ ) → R−π,a (b′ , b∗ ). ∂ ∗ − π(dxd + ax); it is useful to view this map as both D−π,a without the truncation Define Θ−π,a := −x ∂x ′ ∗ ∗ operator, and as −Dπ,a . By definition, R−π,a (b , b ) is the kernel of D−π,a = T runcx ◦ Θ−π,a . Thus, if ξ ∗ ∈ R−π,a (b′ , b∗ ) then Θ−π,a ξ ∗ will consist of only a finite number of positive powers of x. That is, Θ−π,a defines an L(b′ )-linear map Θ−π,a : R−π,a (b′ , b∗ ) → xL(b′ )[x]. We view the right-hand side as a subset of K(b, b′ ). Therefore, by reduction, Θ−π,a induces the mapping Θ−π,a : R−π,a (b′ , b∗ ) → Hπ,a (b′ , b).

P (i) ∗ Notice that Θ−π,a (1) = −Dπ,a (1) = 0 in Hπ,a (b′ , b). Also, writing g−π,i (a, x) = xi + j≥d Bj (a)x−j as in ∗ d−i ∗ the proof of Theorem 2.11, we see that Θ−π,a (g−π,i (a, x)) = −dπx for each i = 1, 2, . . . , d − 1 since g−π,i is in ∗ the kernel of D−π,a and so all negative powers of x vanish, including the constant term. It follows that Θ−π,a induces an L(b′ )-module isomorphism Θ−π,a : R′−π,a (b′ , b∗ ) → Mπ,a (b′ , b). With ξ ∗ ∈ R′−π,a (b′ , b), we have e−π(x

d

+ax)

◦ Φx ◦ eπ(x

d

+ap x)

(ξ ∗ ) = α ¯ ∗−π (a)(ξ ∗ ) + η

where η ∈ L(b′ )[x]. Applying Θ−π,a to both sides we have pe−π(x

d

+ax)

◦ Φx ◦ eπ(x

d

+ap x)

Θ−π,ap (ξ ∗ ) = Θ−π,a α ¯∗−π (a)(ξ ∗ ) + Θ−π,a (η)

which we may rewrite as ¯ ∗−π (a)(ξ ∗ ) − Dπ,ap ◦ απ (a)(η). Θ−π,ap (ξ ∗ ) = p−1 απ (a) ◦ Θ−π,a ◦ α It follows that Θ−π,ap = p−1 α ¯ π (a) ◦ Θ−π,a ◦ α ¯∗−π (a). This is the functional equation. Notice that it relates the conjugate dual Dwork operator Dwork operator α ¯π (a)−1 . 10

(2.5) α ¯ ∗−π (a)

to the inverse

2.6

Frobenius Estimates

Let A(a) = (Aij ) be the matrix of α ¯ (a) with respect to the basis {xi }d−1 i=1 . In this section, we will determine ′ the p-adic order of A as a function of a. Fix b := (p − 1)/p and b := b/p.
Recall from §2.3 that F (a, x) = ij P d−1 r ′ ′ ′ H (a)x ∈ K(b , b ; 0). Thus, ord(H (a)) ≥ b r/d for all ord(a) > − b′ . For i = 1, 2, . . . , d − 1, since r r r≥0 d i ′ ′ ′ i ′ ′ ′ i x ∈ K(b , b ; −ib /d), we see that F (a, x)x ∈ K(b , b ; −ib /d), and so α(a)x ∈ K(b′ , pb′ ; −ib′ /d). From Lemma 2.6, d−1 X Aij (a)xj + Da K(b′ , b) α(a)xi ⊂ j=1

′

for some Aij ∈ L(b′ ; bd (pj − i)). This means ord(Aij (a)) ≥

b′ (pj − i) d

for all ord(a) > −



d−1 d



b′ .

A better estimate than this is often needed for applications. For this, we offer the following exact order for Aij . Notation. Let f (a) be an analytic function over Cp convergent on ord(a) + ρ > 0. We will write f (a) = h(a) + oρ (>) if there is an analytic function h(a) such that ord(f (a)) = ord(h(a)) for all ord(a) + ρ > 0. Theorem 2.13. Suppose d ≥ 2 and p ≥ d + 6. Then there exists ǫ > 0 which depends on p and d such that Aij (a) = where rij :=

π pj−i−(d−1)rij apj−i−drij + oǫ (>) rij !(pj − i − drij )!

 pj−i  . (Note: ǫ → 0+ as p tends to infinity.) d

Remark 2.14. As a consequence, the p-adic absolute values of the entries of the Frobenius A(a) are constant on the unit circle |a| = 1. P⌊r/d⌋ P Proof. Recall from §2.3, we may write F (a, x) = r≥0 Hr (a)xr with Hr (a) := i=0 θi θr−di ar−di ∈ L(b′ ; b′ r/d). In xK(b′ , b), we have X α(a)xi = ψx ◦ F (a, x)xi = Hpj−i xj . j≥1

From Lemma 2.3, we have α(a)xi = µi,1 x + µi,2 x2 + · · · µi,d−1 xd−1 + π(xd + ax)Q, for some Q, where µi,j = Hpj−i − Step 1: Hpj−i =

π pj−i−(d−1)rij pj−i−drij rij !(pj−i−drij )! a

 a r aX Hp[dr+d+j−1−r]−i . (−1)r d d

+ oǫ (>) where rij :=

 pj−i  . d

Recall that θr = π r /r! for r = 0, 1, . . . , p − 1. Consequently, notice that for 1 ≤ i, j ≤ d − 1 and p > d, we have both θr =

πr r!

(2.6)

r≥0

and

θpj−i−dr =

l

p(j−1)−(i−1) d

m

≤r≤

 pj−i  , since d

π pj−i−dr . (pj − i − dr)!

Motivated by this, let us split Hpj−i into two sums:

Hpj−i (a) =

⌉−1 ⌈ p(j−1)−(i−1) d X

θr θpj−i−dr apj−i−dr +

r=0

⌊ pj−i d ⌋ X

r=⌈ p(j−1)−(i−1) ⌉ d

π pj−i−(d−1)r pj−i−dr a . r!(pj − i − dr)!

Step 1 follows after some elementary tedious calculations. Step 2: µij = Hpj−i + oǫ (>). Now,  a r+1 Hp[dr+d+j−1−r]−i ∈ L(b′ ; (b′ /d)[p(dr + d + j − 1 − r) − i] − b′ (r + 1)), d 11

and so, the infinite series in (2.6) is an element in L(b′ ; (b′ /d)(pd − pj − p − i) − b′ ). Comparing this with Hpj−i from Step 1, after tedious calculations, Step 2 follows. Step 3: Aij = µij + oǫ (>). In the notation of Lemma 2.6, µij = η (0) , and so, continuing this notation X Aij − µij = η (r) r≥1

1 > 0. Thus, the series on the right-hand side is an element with η (r) ∈ L(b; (b′ /d)(pj − i) + re) where e := b − p−1 ′ ′ of L(b ; (b /d)(pj − i) + e). Step 3 follows.

3

Fibres

3.1

L-function of the Fibres: L(xd + z¯x, T )

By elementary Dwork theory [12], Dwork’s first splitting function θ(t) := exp π(t − tp ) converges on the closed unit disk D+ (0, 1) and θ(1) is a primitive p-th root of unity in Cp . Let z¯ be an element of a fixed algebraic closure s of Fp and let s := [Fp (¯ z ) : Fp ]. Denote by z the Teichm¨ uller representative in Cp of z¯; notice that z p −1 = 1. Letting fz¯(x) := xd + z¯x, we may define for each n ∈ Z>0 the exponential sum Sn∗ (fz¯) :=

X

xd +¯ zx ¯) T rFpsn /Fp (¯

θ(1)

.

x ¯∈F∗ psn

The associated L-function is L∗ (fz¯, T ) := exp(

X

n≥1

Sn∗ (fz¯)

Tn ). n

Dwork’s splitting function θ(t) defines a p-adic analytic representation of the additive character θ(1)T r(·) . With F (z, x) := θ(xd )θ(zx), by standard Dwork theory, we have (with x the Teichm¨ uller representative of x ¯ in C∗p ) xd +¯ zx ¯) T rFpsn /Fp (¯

θ(1) Therefore,

Sn∗ (fz¯) = x∈Cp

X

ns−1

= F (z, x)F (z p , xp ) · · · F (z p

,xpns−1 =1

ns−1

F (z, x)F (z p , xp ) · · · F (z p

ns−1

, xp

).

ns−1

, xp

).

We now return to our current situation. Let z¯, z, and s be as above. Let b and b′ be real numbers such that 1 ≥ b > p−1 and b/p > b′ > 0. Define the spaces

p−1 p

K(b)z := (K(b′ , b) with the variable a specialized at z) xK(b)z := {h ∈ K(b)z |h is divisible by x}. From Theorem 2.1 and §2.4, Hz := K(b)z /Dz K(b)z

and Mz := xK(b)z /Dz K(b)z

i d−1 are Cp -vector space with bases {xi }d−1 i=0 and {x }i=1 , respectively. Dual to these are the spaces

Rz := (Ra (b′ , b∗ ) with a specialized at z) Rz′ := (R′a (b′ , b∗ ) with a specialized at z). ′ ′ ¯ (a) has coefficients Since, with respect to the basis {xi }d−1 i=0 of both Ha (b , b) and Hap (b , b), the matrix of α ′ in L(b ), we may specialize α ¯ (a) at z. Define α ¯z := α ¯ (a)|a=z . Notice, α ¯ z : Hz → Hzp is an isomorphism from Remark 2.12. s It is well-known that αz is a nuclear operator on the space K(b)z . Thus, since z p −1 = 1, the trace formula for nuclear operators [12, Thm 6.11] tells us that

(ps − 1)T rnuc (αz,s |K(b)z ) = ps T r(¯ αz,s |ker(Dz |K(b)z )) − T r(¯ αz,s |Hz ) 12

where αz,s := αzps−1 ◦ · · · ◦ αzp ◦ αz . On the other hand, Dwork’s trace formula tells us that X s−1 s−1 (ps − 1)T rnuc (αz,s |K(b)z ) = F (z p , xp ) · · · F (z p , xp )F (z, x). xps −1 =1

Observe that the right-hand side is just Dwork’s p-adic analytic representation of the character sum S1∗ (fz¯). That is, (ps − 1)T rnuc (αz,s |K(b)z ) = S1∗ (fz¯). Since αz,sn = (αz,s )n we may use the trace formula to generalize this to

Sn∗ (fz¯) = psn T r(¯ αnz,s |ker(Dz |K(b)z )) − T r(¯ αnz,s |Hz ). Using the well-known identity exp(−

P∞

n

n=0

T r(An ) Tn ) = det(1 − AT ) for finite square matrices A, we obtain

L∗ (fz¯, T ) = exp(

X

Sn∗ (fz¯)

n≥1

=

Tn ) n

det(I − α ¯ z,s T |Hz ) . det(I − ps α ¯ z,s T |ker(Dz |K(b)z )

From §2.4, we know that α∗z,s has 1 as an eigenvector with eigenvalue 1. Thus, det(1 − α∗z,s T |Rz ) = (1 − T )det(1 − α ¯ ∗z,s T |Rz′ ). Also, ker(Dz |K(b)z ) = 0 by Theorem 2.1. Therefore, since det(1 − α ¯ ∗z,s T |Rz′ ) = det(1 − α ¯ z,s T |Mz ), L∗ (fz¯, T ) = (1 − T )det(1 − α ¯ z,s T |Mz ) = (1 − T ) Equivalently, for every n ≥ 1,

d−1 Y i=1

(1 − πi (¯ z )T ) ∈ Z[ζp ][T ].

−Sn∗ (fz¯) = 1 + π1 (¯ z )n + · · · + πd−1 (¯ z )n .

Now, had we used x¯ = 0 in the definition of the character sum Sn∗ (fz¯), that is, X

Sn (fz¯) :=

xd +¯ zx ¯) T rFpsn /Fp (¯

θ(1)

,

x ¯∈Fpsn

then since Sn (fz¯) = 1 + Sn∗ (fz¯), we have L(fz¯, T ) := exp(

X

Sn (fz¯)

n≥1

d−1 Y Tn (1 − πi (¯ z )T ). ) = det(I − α ¯ z,s T |Mz ) = n i=1

Notice that, from the relative functional equation (2.5), there is a nonzero constant c, dependent on p, d, and deg(¯ z ), such that cT d−1 L(fz¯, p−s T −1 ) = L(fz¯, T ), where the bar on the left means complex conjugation. Note, when d is odd, since −(xd + ax) = (−x)d + a(−x), the L-function has real coefficients since the exponential sums are in this case real. Newton Polygon. Let us determine the Newton polygon of the L-function L(fz¯, T ). For simplicity, we will ′ ′ p−1 ˜ . Consider the assume z¯ ∈ F∗p . Set b := p−1 p and b := b/p. Let ξ ∈ Cp such that ordp (ξ) = b /d, and let ξ := ξ i i d−1 ′ basis {ξ x }i=1 for the space Ma (b , b). From the beginning of §2.6, we have i i α(a)ξ ¯ x =

d−1 X j=1

(Ai,j ξ i−j )ξ j xj ∈ Map (b′ , pb′ ; 0)

13

Thus, with respect to this basis, the matrix A(a) = (Aij ) of α ¯ (a), acting on the right, may be written as 

  A(a) =  

˜ 0,1 ξA ˜ ξA1,0 .. . ˜ d−1,1 ξA

A0,0 A1,0 .. . Ad−1,0

ξ d−1 A0,d−1 ξ d−1 A1,d−1 .. . ξ˜d−1 Ad−1,d−1

··· ··· .. . ···

    

P m where ξ˜j Ai,j = Aij ξ i−j and ord(Ai,j ) ≥ 0. Now, det(1 − A(a)T ) = d−1 where m=0 cm T      X X cm := (−1)m sgn(σ) ξ˜u1 Aσ(u1 ),u1 ξ˜u2 Aσ(u2 ),u2 · · · ξ˜um Aσ(um ),um 1≤u1 1, the (local) fundamental solution matrix takes the form Y(a) = a−1/2 V (a)S(a) where v1 (a) := v(1/a) and v2 (a) := v¯(1/a), and   v1 (a) v2 (a) V (a) := (− 12 a−2 + 3a2 κπ)v1 + a−1 v1′ (a) (− 12 a−2 − 3a2 κπ)v2 + a−1 v2′ (a) 19

and

3

S(a) :=

eκπa 0

0 3

e−κπa  m1 From the previous section, there is a constant matrix M := m3

!

.

m2 m4



such that Aa2 Yφ = YM . That is,

Aa2 a−p/2 V (ap )S(ap ) = a−1/2 V (a)S(a)M.

(4.7)

b of analytic functions on 1 < |a| < e, e > 1 unspecified, adjoined the algeLet us view this equation over the ring R √ 3p 3p 3 3 κπa3 −κπa3 b braic (over R) functions a, e ,e , eκπa , e−κπa satisfying the compatibility relations: eκπa e−κπa = 1, 3p 3p 3 3p p 3p 3 3p p 3p eκπa e−κπa = 1, eκπa e−κπa = θ(κa3 )eπ(κ −κ)a , and eκπa eκπa = θ(κa3 )eπ(κ +κ)a , where θ is Dwork’s first splitting function on Fp . Using the compatibility relations, we may write ! p 3p p 3p m1 θ(κa3 )eπ(κ −κ)a m2 θ(κa3 )eπ(κ +κ)a p −(p−1)/2 −1 ∗ . = V (a) Aa2 V (a )a p 3p p 3p m2 θ(−κa3 )e−π(κ +κ)a m4 θ(−κa3 )e−π(κ −κ)a b thus the ride-hand side must be as well. It follows from Lemma 4.3 Notice that the left-hand side belongs to R, below that: Theorem 4.2. If p ≡ 1, 7 mod(12), then

M=



m1 0

0 m4



,

M=



0 m3

m2 0



.

else if p ≡ 5, 11 mod(12), then

Lemma 4.3. For p ≡ 1, 7 mod(12), we have ordp (κp − κ) > 0 and ordp (κp + κ) = 0. Otherwise, for p ≡ 5, 11 mod(12), we have ordp (κp − κ) = 0 and ordp (κp + κ) > 0 Proof. Recall, κ :=

2i √ . 3 3

Thus,  h p  i 2 2 i √ √ − if p ≡ 1 mod(4) 3 3  3 3 i h κp − κ = p 2 2 −i √ + 3√3 if p ≡ 3 mod(4). 3 3

Let us concentrate on p ≡ 1 mod(4). By Fermat’s little theorem, 2p = 2 + pOCp and 3p = 3 + pOCp , where OCp = {c ∈ Cp : |c| ≤ 1}. Consequently, we have √ 6 3i κp − κ = √ (1 − 3(p−1)/2 ) + pOCp . (3 3)p   Thus, κp − κ has positive p-adic order if and only if 3(p−1)/2 ≡ 1 mod(p). By Euler’s criterion, 3(p−1)/2 ≡ p3
mod(p), where the right-hand side is the Legendre symbol. Since p3 = 1 if and only if p ≡ 1 mod(3), the cases p ≡ 1, 5 mod(12) for κp − κ follow from quadratic reciprocity. The other cases are similar.   0 1 Notice that v1 (−a) = v2 (a) and so v1 (−a)′ = −v2 (a). Thus, if T := = T −1 , then V (−a)T = V (a). 1 0 Next, since the change a 7→ −a does not affect Aa2 , changing a 7→ −a in (4.7) gives V (a)−1 Aa2 V (ap )(−1)(p−1)/2 a−(p−1)/2 = S(a)T M T S(ap). Thus, (−1)(p−1)/2 T M = M T which demonstrates that if p ≡ 1 mod(12), then m1 = m4 , else if p ≡ 7 mod(12), then −m1 = m4 . Similarly, if p ≡ 5 mod(12), then m2 = m3 , else if p ≡ 11 mod(12) then −m2 = m3 . Next, since det(Y) is locally constant, from Aa2 Yφ = YM we see that det(M ) = det(Aa2 ), and this equals p if p ≡ 1 mod(3) and g := −g2 ((p2 − 1)/3) if p ≡ −1 mod(3). Putting everything together, we have:   √ p 0 √ if p ≡ 1, 7 mod(12) M= 0 ± p 20

(of course,

√ p may not be the positive square root) and   √ −g √0 M= 0 ± −g

if p ≡ 5, 11 mod(12)

where “±” means positive if p ≡ 1, 5 mod(12) and negative if p ≡ 7, 11 mod(12). Symmetric powers at infinity. As demonstrated above, {y1 , y2 } forms a formal basis for the solution space d2 d 4 (around a point near infinity) of the differential operator ˜l := da We define the k-th 2 − (1/a) da + 4πa /3. ˜ ˜ symmetric power of l, denoted by lk , as the unique monic differential equation of order k + 1 that annihilates every 3 zj := y1k−j y2j = a−k/2 e(k−2j)κπa v1 (a)k−j v2 (a)j for j = 0, 1, . . . , k. Define

b := analytic functions on 1 < |a| < e, e unspecified. R

b Let Y (a) ∈ R b be a solution of ˜lk which converges We are interested in studying the solutions of ˜ lk over the ring R. k ˜ b then locally on 1 < |a| < e. Fix 1 < |z| < e. Since {zj }j=0 is a formal basis of lk , if Y is a solution of ˜lk over R, − (in D (z, 1)) using Remark 4.1 we have the relation Y (a) =

k X

3

cj a−k/2 v j (1/a)¯ v k−j (1/a)[eκπ(a−z) ]2j−k

j=0

for some constants cj ∈ Cp . Let us rewrite this as a

k/2

Y (a) =

p−1 X

3

Sr (a)[eκπ(a−z) ]r

(4.8)

r=0

where for each r = 0, 1, . . . , p − 1, Sr (a) :=

X

3 (2j−k)−r p

cj v j (1/a)¯ v k−j (1/a)[epκπ(a−z) ]

j∈Er

with Er := {j ∈ Z ∩ [0, k] : (2j − k) ≡ r mod(p)}.

3

Since v(1/a) and v¯(1/a) converge on D+ (z, 1) ⊂ D− (∞, 1), and e±pκπ(a−z) and Y (a) converge on D+ (z, 1), equation (4.8) √ shows us that the functions exp(rκπ(a − z)3 ), for r = 0, 1, . . . , p − 1, are linearly dependent + over M (z, 1)[ a], where M+ (z, 1) is the field of meromorphic functions on the closed unit disc around √ z. However, Lemma 4.4 below shows us that these functions are actually linearly independent over M+ (z, 1)[ a]. Consequently, we must have Sr = 0 for each r ≥ 1, and so, X cj v j (1/a)¯ v k−j (1/a) exp[(2j − k)κπ(a − z)3 ]. ak/2 Y (a) = S0 (a) = (2j−k)≡0 mod(p)

Notice that the right-hand side now converges on D+ (z, 1). Thus, if k is odd, since Y (a) converges on D+ (z, 1), we see that ak/2 ∈ M+ (z, 1). However, by Lemma 4.5 below, this cannot be. Therefore, for k odd we must have Y = 0. b else if k is odd then ker(˜lk |R) b = 0. Thus, for k even {zj : 0 ≤ j ≤ k, p|(k − 2j)} is a basis for ker(˜lk |R), 2

2

d π a Symmetric powers of the Airy operator. (cf. [13, §7]) Define the Airy differential operator l := da 2 + 3 b and denote by lk its k-th symmetric power. We wish to study ker(lk |R). b = 0 since any solution Y ∈ R b of lk produces a solution Y (a2 ) to ˜lk . Assume 2|k. Notice For k odd, ker(lk |R) that v1 (−a) = v2 (a). Consequently, for 0 ≤ j ≤ k/2 and p|(k − 2j), 3

3

3

3

w ej+ (a) := e(k−2j)κπa v1 (a)k−j v2 (a)j + e−(k−2j)κπa v1 (a)j v2 (a)k−j

is an even function while

w ej− (a) := e(k−2j)κπa v1 (a)k−j v2 (a)j − e−(k−2j)κπa v1 (a)j v2 (a)k−j 21

b is an odd function. Clearly, {a−k/2 w ej+ , a−k/2 w ej− |0 ≤ j ≤ k/2, p|(k − 2j)} is a basis for ker(˜lk |R). −k/2 + From this, if 4|k then {a w ej (a)|0 ≤ j ≤ k/2, p|(k −2j)} is a set of even functions whereas {a−k/2 w ej− (a)|0 ≤ + + b = span{w (a)|0 ≤ j ≤ k/2, p|(k − 2j)} where w is defined by j ≤ k/2, p|(k − 2j)} are odd. Thus ker(lk |R) j j + 2 −k/2 + b = span{w− (a)|0 ≤ j < k/2, p|(k − 2j)}, where wj (a ) = a w ej (a). Similarly, if 2|k but 4 ∤ k, then ker(lk |R) j wj− is defined by wj− (a2 ) = a−k/2 w ej− (a). Thus, j k  k  1 + if 4|k  j k 2p b k dimCp ker(lk |R) = if 2|k but 4 ∤ k 2p    0 if k odd. Appendix. Lemma 4.4. Let M+ (0, 1) denote the field of meromorphic functions on D+ (0, 1) in the variable a. Then the √ p pπa + + polynomial X − e ∈ M (0, 1)[X] is irreducible over M (0, 1)[ a]. √ Proof. It is clear that X p − epπa is either irreducible or completely reducible over M+ (0, 1)[ a]. If the latter, then we would have √ eπa = H1 (a) + aH2 (a) for some H1 , H2 ∈ M+ (0, 1). This implies that eπa satisfies a quadratic polynomial whose coefficients lie in M+ (0, 1). Thus, X p − epπa is completely reducible over M+ (0, 1). In particular, eπa ∈ M+ (0, 1). However, this is not true by the following lemma. Lemma 4.5. (a) exp(πa) ∈ / M+ (0, 1). (b) Let a1/2 be a branch of square root centered around z ∈ C∗p . Then a1/2 ∈ / M+ (z, 1). Proof. We know that exp(πa) converges on D− (0, 1) and is divergent on the boundary |a| = 1. Suppose there exists analytic functions f (a) and g(a) on D+ (0, 1) such that we have the equality eπa = f (a)/g(a) whenever |a| < 1. Raising this to the p-th power gives the equality epπa = f (a)p /g(a)p for all |a| ≤ 1. Since exp(pπa) has no poles, g(a) must be invertible on D+ (0, 1). Let h(a) := 1/g(a) and note this converges on D+ (0, 1). Then exp(πa) = f (a)h(a) which shows we may extend exp(πa) to an analytic function on D+ (0, 1). Contradiction. This proves (a). A similar argument proves (b).

5

Cohomological Formula for Mk (T )

Define Mk∗ (T ) :=

k Y Y

λ∈|Gm | i=0

(1 − π1 (λ)i π2 (λ)k−i T deg(λ) )−1

where |Gm | denotes the set of Zariski closed points of Gm := P1Fp \ {0, ∞}. In this section, we will define a Frobenius operator β¯k on the (symmetric power) cohomology spaces Hk0 and Hk1 such that Mk∗ (T ) =

det(1 − β¯k T |Hk1 ) . det(1 − pβ¯k T |Hk0 )

From this, we will deduce that Mk (T ) (and Mk∗ (T )) is a rational function. We will also find P its fields of definition. The relationship between Mk (T ) and Mk∗ (T ) is as follows. If we write Mk (T ) = exp s≥1 Ns Ts , then taking the logarithmic derivative of the definition of Mk (T ) shows Ns =

k X X

s

deg(¯ a)[π1 (¯ a)k−j π2 (¯ a)j ] deg(¯a) .

a ¯∈Fps j=0

Using the same procedure, we may calculate Ns∗ , where Mk∗ (T ) = exp Ns =

k X

P

s≥1

(π1 (0)k−j π2 (0)j )s + Ns∗ .

j=0

22

s

Ns∗ Ts . It follows that (5.1)

d + πax on K(b′ , b), and notice that, formally Differential Operator. Define the operator ∂a := a da

∂a = e−π(x

3

+ax)

◦a

3 d ◦ eπ(x +ax) . da

It follows that ∂a and Da commute, and so ∂a defines an operator on both Ha (b′ , b) and Ma (b′ , b). (k) Denote by Ma (b′ , b) the k-th symmetric power of Ma (b′ , b) over L(b′ ). It is easy to see that ∂a induces an (k) endomorphism on Ma (b′ , b) by extending linearly k X

∂a (u1 · · · uk ) :=

i=1

u1 · · · u ˆi · · · uk ∂a (ui )

where u ˆi means we are leaving it out of the product. Matrix representation of ∂a . With the L(b′ )-basis {v, w} of Ma (b′ , b), where v := x and w := x2 , fix the (k) ′ ordered basis (v k , v k−1 w, . . . , wk ) of Ma (b′ , b). Since Da (1) = 3πx3 + πax, we see that x3 ≡ − ax 3 in Ma (b , b). −πa2 Therefore, ∂a (v) = πaw and ∂a (w) = 3 v. For each j = 0, 1, . . . , k, we have ∂a (v j wk−j ) = jv j−1 ∂a (v)wk−j + (k − j)v j wk−j−1 ∂a (w) = jπav j−1 wk−j+1 + (k − j)

−πa2 j+1 k−j−1 v w . 3

d − Gk where Thus, ∂a , acting on row vectors, has the matrix representation a da



   Gk :=    

0

kπa 0

2

−πa 3

(k − 1)πa

2

−2πa 3

..

..

.

. −kπa2 3



   .   πa  0

1 ′ Dwork Operator. For p−1 ¯ (a) is an isomorphism of Ma (b′ , b) onto Map (b′ , b) p ≥ b > p−1 and b/p ≥ b > 0, α d ¯ (a) = α ¯ (a)∂a (since a da satisfying ∂ap α and ψx commute with no p-factor). Thus α(a) ¯ induces an isomorphism (k) ′ (k) ′ (k) α ¯ k (a) : Ma (b , b) → Map (b , b) defined by acting on the image of (u1 , . . . , uk ) ∈ Ma (b′ , b)k in Ma (b′ , b) by

α ¯ k (a)(u1 · · · uk ) := α(a)(u ¯ ¯ (a)(uk ). 1) · · · α ¯ k (a) = α ¯ k (a)∂a . Note, ∂ap α Define the operator ψa : L(b′ ) → L(pb′ ) by ψa :

∞ X i=0

(k) Ma (b′ , b)

We may extend this map to (k) (k) ψa : Map (b′ , b) → Ma (pb′ , b) by

i

Bi a 7−→

∞ X

Bpi ai .

i=0

(k)

as follows. Fix the basis {v i wk−i }ki=0 of Ma (b′ , b). Define the map

k X i=0

hi v i wk−i 7−→

k X

ψa (hi )v i wk−i .

i=0

It is not hard to show that ψa ∂ap = p∂a ψa . (k) (k) (k) Denote by Ma (b) the space Ma (b, b). We define the Dwork operator βk := ψa ◦ α ¯ k (a) on Ma (b) by noting (k)

α ¯ k (a)

ψa

(k)

(k)

Ma (b) −−−−→ Ma (b/p, b) −−−−→ Ma (b). (k)

This is a nuclear operator on Ma (b) in the sense of [14]. Cohomology. Define the cohomology spaces Hk0 := ker(∂a |M(k) a (b))

(k) Hk1 := M(k) a (b)/∂a Ma (b).

23

Since ¯ k (a) = p∂a ◦ ψa ◦ α ¯ k (a) = ∂a ◦ pβk , βk ◦ ∂a = ψa ◦ α ¯ k (a) ◦ ∂a = ψa ◦ ∂ap ◦ α βk induces endomorphisms

β¯k : Hk0 → Hk0

and β¯k : Hk1 → Hk1 .

These maps are in fact isomorphisms; this follows by defining a right-inverse to βk as in Remark 2.12. Dwork Trace Formula. From the following exact sequence and chain map: (k)

∂

(k)

(k)

∂

(k)

0 −−−−→ Hk0 −−−−→ Ma (b) −−−a−→ Ma (b) −−−−→       β s  ps βks y ps β¯ks y y k y

Hk1 −−−−→  β¯s y k

0   y

0 −−−−→ Hk0 −−−−→ Ma (b) −−−a−→ Ma (b) −−−−→ Hk1 −−−−→ 0

it follows from [14, Proposition 8.3.10] that the alternating sum of the traces is zero: s (k) ¯s 1 T r(ps β¯ks |Hk0 ) − T rnuc (ps βks |M(k) a (b)) + T rnuc (βk |Ma (b)) − T r(βk |Hk ) = 0.

Hence,

s ¯s 1 ¯s 0 (ps − 1)T rnuc (βks |M(k) a (b)) = p T r(βk |Hk ) − T r(βk |Hk ).

(5.2)

Next, we have Dwork’s trace formula (ps − 1)T rnuc (βks |M(k) a (b)) = where

s−1

α ¯ k (a; s) := α ¯ k (ap

X

ps −1 =1 a∈C∗ p ,a

s−2

)◦α ¯ k (ap

T rM(k) (b) α ¯ k (a; s)

(5.3)

a

) ◦ ···◦ α ¯ k (a) ∗

uller representative in Cp , Cohomological Formula for Mk (T ). Recall from §3.1, if z¯ ∈ Fp and z is its Teichm¨ then α ¯ z was defined to be the operator α ¯ (a) specialized at a = z. Using this, define s−1

α ¯ (k) ¯ k (ap z,s := α

s−2

)◦α ¯k (ap

) ◦ ···◦ α ¯ k (a)|a=z .

Using (5.2) and (5.3) for the first two equalities (resp.) of the following, we have X det(1 − β¯k T |Hk1 ) Ts s s (k) = exp (p − 1)T r (β |M (b)) nuc k a s det(1 − pβ¯k T |Hk0 ) s≥1 = exp

X

X

s≥1 a∈Cp ,aps −1 =1

= exp

X

T rM(k) (b) (¯ αk (a; s)) a

X

X

T rM(k) (b) (¯ αk (a; rm))

X

X

α(k) T rCp (¯ a,rm )

X

X

∗ r≥1 a a)=r m≥1 ¯ ∈F ,deg(¯

= exp

X

p

a=T eich(¯ a)

a

∗ r≥1 a ¯ ∈F ,deg(¯ a)=r m≥1

= exp

X

p

a=T eich(¯ a)

=

Y

r≥1

a=T eich(¯ a)

Y

∗ a ¯ ∈Fp ,deg(¯ a)=r

T rm rm

T rm rm

(k) m αa,r ) ) T rCp ((¯

∗ r≥1 a ¯ ∈F ,deg(¯ a)=r m≥1 p

Ts s

T rm rm

r −1/r ¯ (k) . [detCp (1 − α a,r T )]

(5.4)

a=T eich(¯ a)

(k)

From §3.1 we know that α ¯ a,r has two eigenvalues π1 (a) and π2 (a) on Ma . Since α ¯ a,r is the k-th symmetric power of α ¯ a,r , its eigenvalues are π1 (a)k−j π2 (a)j for j = 0, 1, . . . , k. Thus, det(1 − α ¯ (k) a,r T ) =

k Y

j=0

(1 − π1 (a)k−j π2 (a)j T ). 24

This means (5.4) equals Y

r≥1

Y

∗ a ¯ ∈Fp ,deg(¯ a)=r

a=T eich(¯ a)

k Y

j=0

(1 − π1 (a)k−j π2 (a)j T r )−1/r

which is precisely Mk∗ (T ). Theorem 5.1. Mk (T ) is a rational function with coefficients in Z[ζp ]. If p ≡ 1 mod(3), then the coefficients of Mk (T ) lie in the ring of integers of the unique subfield L of Q(ζp ) with [L : Q] = 3. However, if p ≡ −1 mod(3) then Mk (T ) has integer coefficients. Proof. We will first prove rationality by using a result P of Borel-Dwork. From the cohomological description, s Mk (T ) is p-adic meromorphic. Next, write Mk (T ) = exp s≥1 Ns Ts as in the introduction, with Ns =

k X X

s

deg(¯ a)[π1 (¯ a)k−j π2 (¯ a)j ] deg(¯a) .

a ¯∈Fps j=0

Since |πj (¯ a)|C ≤ pdeg(¯a)/2 Riemann Hypothesis for curves, |Ns |C ≤ csp(k+1)s/2 for a positive integer c Pby the Ns s independent of s. Thus s≥1 s T has at least a p−(k+1)/2 radius of convergence on C. Consequently, Mk (T ) has a positive C-radius of convergence. Let ζp is a primitive p-th root of unity. Since L(x3 + a ¯x, T ) is a polynomial with entries in Z[ζp ], we see that the roots live in Z[ζp ]-rational orbits under the action of the Galois group Gal(Q/Q(ζp )). Hence, the symmetric Qk polynomial j=0 (1 − π1 (¯ a)k−j π2 (¯ a)j T ) must also have coefficients in Z[ζp ]. It follows that Mk (T ) has coefficients in Z[ζp ]. Since Mk (T ) is p-adic meromorphic and converges on a disc of positive radius in C, and the coefficients lie in a fixed number field, by the Borel-Dwork theorem [4, Theorem 3] Mk (T ) is a rational function with coefficients in Z[ζp ]. Next, with each λ ∈ F∗p we may define σλ ∈ Gal(Q(ζp )/Q), where , by σλ (ζp ) := ζpλ ; note, every element of Gal(Q(ζp )/Q) may be represented uniquely by such a σλ . Now, X T rF k /Fp (¯x3 +λ2 a¯x¯) X T rF k /Fp [λ3 (¯x3 +¯ax¯)] ζp p ζp p = x ¯∈Fpk

x ¯∈Fpk



= σλ3 

X

x ¯∈Fpk

This means σλ3 (L(x3 + a ¯x, T )) = L(x3 + λ2 a ¯x, T ). Consequently,



x3 +¯ ax ¯) T rF k /Fp (¯ . ζp p

σλ3 (Mk (T )) = Mk (T ), 3

and so Mk (T ) is fixed by the subgroup {λ |λ ∈ F∗p }. Since this subgroup has index r := gcd(3, p − 1) in the Galois group, the coefficients of Mk (T ) must lie in the ring of integers of the unique subfield of Q(ζp ) of degree r over Q.

6 6.1

Cohomology of Symmetric Powers Hk0 for all odd k, or for all k < 2p even

Recall, with respect to the basis {x, x2 } of Ma (b′ , b), ∂a has the matrix representation (acting on row vectors):   0 πa d − . ∂a = a −πa2 0 da 3 Thus, if (C1 , C2 ) is a solution of this differential equation then C2 must satisfy the Airy differential equation: 2 y ′′ + π3a y = 0. (k)

Theorem 6.1. Let k be a positive integer. If k is odd, or k is even with k < 2p, then Hk0 := ker(∂a |Ma (b′ , b)) = 0. (k)

Proof. Every element of Ma (b′ , b) which is annihilated by ∂a corresponds to a unique solution in L(b′ ), and b of the scalar equation lk , where lk was defined in §4.1. By the methods of §4.1, if k is odd, then no such hence R, b If 2|k but 4 ∤ k, and k < 2p, then again, no such solution exists. If 4|k, then one solution of solution exists in R. b however, this corresponds to v1 (a)k/2 v2 (a)k/2 which is clearly not an element of L(b′ ). lk exists in R, Conjecture 6.2. Hk0 = 0 for all positive integers k.

25

6.2

Hk1 for odd k < p

We will assume k is odd and k < p throughout this section. The two vectors v := x and w := x2 form an (k) L(b)-basis of Ma (b). Thus, the set {v j wk−j |0 ≤ j ≤ k} is an L(b)-basis of Ma (b). With δj := bj 3 , define M(k) a (b; δ + ρ) :=

k M

L(b; δj+1 + ρ)v k−j wj

j=0

= L(b; δ1 + ρ) ⊕ L(b; δ2 + ρ) ⊕ · · · ⊕ L(b; δk+1 + ρ) and M(k) a (b) :=

[

ρ∈R

M(k) a (b; δ + ρ).

(k)

We will write Ma (b; ρ) with the δ implied. Define (k−1)/2

M

Vk := Ck+1 ⊕ p

(Cp a)v k−2j w2j

(6.1)

j=0

Vk (b; ρ) := Vk ∩ M(k) a (b; ρ) [ Vk (b) := Vk (b; ρ). ρ∈R

1 Theorem 6.3. Let p ≥ 5 be a prime. Suppose k is odd and k < p. Let b be a real number such that e := b − p−1 > 0. Then we have the following decomposition: (k) M(k) a (b) = Vk (b) ⊕ ∂a Ma (b). (k)

Furthermore, ∂a is injective on Ma (b). (k)

It will be useful to recall, if we order the basis of Ma (b) as (v k , v k−1 w, . . . , wk ), then ∂a has the matrix d representation ∂a = a da − Gk , acting on row vectors, where 

0 2

   Gk :=    

−πa 3

kπa 0 −2πa2 3

(k − 1)πa ..

.

. −kπa2 3

That is, if we write the row vector ξ = (ξ1 , . . . , ξk+1 ) for ∂a ξ (k)

..



Pk=1 j=1

      πa  0

ξj v k−j wj then

means (ξ1 , . . . , ξk+1 )(a

d − Gk ). da

(k)

Lemma 6.4. Ma (b; 0) ⊂ Vk (b; 0) + Ma (b; e)Gk . Proof. Odd Column Case. For n ≥ 2, consider the vector (0, . . . , Bn an , . . . , 0) where an is in the (2j + 1)-entry with all others zero, and Bn an ∈ L(b; δ2j+1 ). Suppose n ≤ (k + 1)/2 − j. Then 2j+1 entry

where

2(n−1+j)+1 entry z }| { z}|{ n ηj,n , . . . , 0) (0, . . . , Bn a , . . . , 0) = (0, 0, . . . , 0, C2(j+1) , 0, C2(j+2) , · · · )Gk + (0, . . . ,

ηj,n :=

n−2 Y l=0

and for i = 1, 2, . . . , n − 1, we have C2(j+i)

−3i = π

3[k − (2(j + l) + 1)] Bn a ∈ L(b; δ2(n−1+j)+1 ) [2(j + l) + 1]

Qi−2

l=0 [k − (2(j + l) + 1)] Qi−1 l=0 [2(j + l) + 1]

26

!

Bn an−1−i ∈ L(b; δ2(j+i) + e)

Next, suppose n ≥ (k + 1)/2 − j + 1. Then 2j+1 entry

z }| { (0, . . . , Bn an , . . . , 0) = (0, 0, . . . , 0, C2(j+1) , 0, C2(j+2) , · · · )Gk

where, for m = 1, 2 . . . , (k + 1)/2 − j,

C2(j+m) :=

(−3)m (j − k−1 2 )m−1 Bn an−m−1 . 1 2π(j + 2 )m

Notice that, since k < p, the constants in front of the series are p-adic units. P∞ i }. Let g(a) := Even Column Case. Fix j ∈ {1, . . . , k+1 i=0 Bi a ∈ L(b; δ2j ). Then we may write 2 entry 2j

entry 2j

z}|{ z}|{ (0, . . . , 0, g(a) , 0, . . . , 0) = ξGk + (0, . . . , 0, B0 , 0, . . . , 0)

where

ξ := (ξ1 , 0, ξ3 , 0, . . . , ξ2j−1 , 0, 0, . . . , 0) where, for m = 0, 1, . . . , j − 1, ξ2j−1−2m :=

∞ X (1 − j)m Bi+1 ai+m ∈ L(b; δ2j−1−2m + e). 2(−3)m π( k2 − j + 1)m+1 i=0

  (k) Lemma 6.5. Vk (b) ∩ Ma (b)Gk = {0}.

  (k) (k) Proof. We will proceed by induction. Let η = (η1 , . . . , ηk+1 ) ∈ Vk ∩ Ma (b)Gk . Then there is a ξ ∈ Ma (b)

such that ξGk = η. In particular, η1 = c1 and c2 we may write

−πa2 3 ξ2 .

Hence, ξ2 = 0. Suppose ξ2(j−1) = 0, then for some p-adic units

η2j−1 = c1 aξ2(j−1) + c2 a2 ξ2j

which shows ξ2j = 0. Thus, ξ2j = 0 for j = 1, 2, . . . , k+1 2 . Next, since ηk+1 = πaξk we see that ξk = 0. Suppose ξ2j+1 = 0, then since there are units c1 and c2 such that η2j = c1 aξ2j−1 + c2 a2 ξ2j+1 we see that ξ2j−1 = 0. (k)

(k)

(k)

Lemma 6.6. Let u ∈ Ma (b) such that uGk ∈ Ma (b; ρ). Then u ∈ Ma (b; ρ + e). Consequently, Gk is an (k) injective operator on Ma (b). 2 Proof. We will proceed by induction. Let ξ := uGk . Then ξ1 = −π 3 a u2 ∈ L(b; δ1 ). Thus, u2 ∈ L(b; δ2 + e) as desired. Next, suppose u2j ∈ L(b; δ2j + e). Then πau2j ∈ L(b; δ2j+1 ). From ξ = uGk , there are p-adic units c1 and c2 such that c2 πa2 u2j+2 ∈ L(b; δ2j+1 ). ξ2j+1 = c1 πau2j − 3 2

Hence, c2 πa 3 u2j+2 ∈ L(b; δ2j+1 ), and this implies u2j+2 ∈ L(b; δ2j+2 + e). Next, by hypothesis, ξk+1 = πauk ∈ L(b; δk+1 ) and so, uk ∈ L(b; δk + e). Suppose u2j+1 ∈ L(b; δ2j+1 + e). Then πa2 u2j+1 ∈ L(b; δ2j ). As before, from ξ = uGk , there are p-adic units c1 and c2 such that ξ2j = c1 πau2j−1 − c2 πa2 u2j+1 ∈ L(b; δ2j ). As before, it follows that u2j−1 ∈ L(b; δ2j−1 + e). (k)

(k)

Lemma 6.7. Ma (b; 0) ⊂ Vk (b; 0) + ∂a Ma (b; e).

27

(k)

(k)

Proof. Let u ∈ Ma (b; 0). Set u(0) := u. We know there exists unique η (0) ∈ Vk (b; 0) and ξ (0) ∈ Ma (b; e) such that u(0) = η (0) + ξ (0) Gk . d Set E := a da . Then

u(1) : = u(0) − η (0) + ∂a ξ (0)

= Eξ (0) ∈ M(k) a (b; e). (k)

Again, there exists unique η (1) ∈ Vk (b; e) and ξ (1) ∈ Ma (b; 2e) such that u(1) = η (1) + ξ (1) Gk . Set u(2) := u(1) − η (1) + ∂a ξ (1) . Continue this procedure h-times: u(h) := u(h−1) − η (h−1) + ∂a ξ (h−1) ∈ M(k) a (b; he). Adding these together produces u(h) = u(0) − (k)

h−1 X

η (i) + ∂a

h−1 X

ξ (i) .

i=0

i=0

Letting h → ∞ we see that u(h) → 0 in Ma (b). Thus, u(0) =

∞ X i=0

η (i) − ∂a

∞ X i=0

ξ (i) ∈ Vk (b) + ∂a M(k) a (b).

(k)

(k)

(k)

Lemma 6.8. Let u ∈ Ma (b) be such that ∂a u ∈ Ma (b; ρ). Then u ∈ Ma (b; ρ + e). (k)

(k)

Proof. Suppose u 6= 0. Choose c ∈ R such that u ∈ Ma (b; c) but u ∈ / Ma (b; c + e). By hypothesis, ∂a u ∈ (k) Ma (b; ρ). Thus (k) (k) −uGk = ∂a u − Eu ∈ M(k) a (b; ρ) + Ma (b; c) = Ma (b; l). (k)

where l := min{ρ, c}. Hence, u ∈ Ma (b; l + e) which means l = ρ. (k)

Corollary 6.9. ∂a is an injective operator on Ma (b). (k)

Lemma 6.10. Vk (b) ∩ ∂a Ma (b) = {0}. (k)

(k)

Proof. Let u ∈ Vk (b) ∩ ∂a Ma (b). Assume u 6= 0. Then, we may find ρ ∈ R such that u ∈ Ma (b; ρ) but (k) (k) (k) u∈ / Ma (b; ρ + e). By hypothesis, there exists η ∈ Ma (b) such that ∂a η = u. Thus, η ∈ Ma (b; ρ + e). Now, u + ηGk = Eη ∈ M(k) a (b; ρ + e) (k)

which means that we may find ζ ∈ Vk (b; ρ + e) and w ∈ Ma (b; ρ + 2e) such that Eη = ζ + wGk . Therefore, u = ζ + (w − η)Gk . Since u ∈ Vk (b), by uniqueness, u = ζ (and η = w). However, this means (k) u ∈ Vk (b; ρ + e) ⊂ Ma (b; ρ + e) which is a contradiction. This concludes the proof of Theorem 6.3.

28

6.3

Hk1 for even k < p

In this section, while we attempt to proceed in a similar manner to the last, we are only able to prove a partial decomposition theorem. The crucial difference is that when k is even, Gk has a kernel. Assume k is even and k < p. With δj := bj 3 , define M(k) a (b; δ + ρ) :=

k M

L(b; ρ + δj+1 )v k−j wj

j=0

(k)

As we did in the previous section, we will write Ma (b; ρ) with the δ implied. Ordering the basis as (v k , v k−1 w, . . . , wk ), fa(k) (b; ρ) ⊂ M(k) define the subspace M a (b; ρ) as f(k) (b; 0) := L(b; δ1 ) ⊕ L(b; δ2 ) ⊕ · · · ⊕ L(b; δk ) ⊕ {0} M a

Next, define

k/2−1

Vk := Ck+1 ⊕ p and the vector

M

Cp av k−2j w2j ,

j=0

k := (K1 , 0, K3 , 0, . . . , 0, Kk+1 ) where for j = 0, 1, . . . , k/2 K2j+1 :=

Qj

k i=1 [ 2

− (i − 1)] k/2−j a . j! 3k/2−j

Notice that the kernel of Gk is L(b)k. Define Iker (b; ρ) = aL(b)k ∩ M(k) a (b; ρ) and Iker (b) := aL(b)k. Theorem 6.11. Let p ≥ 5. Suppose k is even and k < p. Let b be a real number such that b −

1 p−1

> 0. Then

f(k) M(k) a (b) = Vk (b) ⊕ ∂a Ma (b) ⊕ Iker (b).

The proof will consist of several lemmas. Lemma 6.12. Set e := b −

1 p−1

> 0. Then

f(k) M(k) a (b; 0) ⊂ Vk (b; 0) + Ma (b; e)Gk + Iker (b; 0).

The proof of this is a bit involved.

Even Entry Case. Consider (0, . . . , g(a), . . . , 0) where g(a) = other entries are zero. Then

P

i≥0

Bi ai ∈ L(b; δ2j ) is in the 2j-entry, and all

2j entry

where

2j entry

z}|{ z}|{ (0, . . . , g(a) , . . . , 0) = (C1 , 0, C3 , 0, . . . , C2j−1 , 0, 0, . . . , 0)Gk + (0, . . . , B0 , . . . , 0) C2j−2l−1 =

Ql−1

X m=0 (j − l + m) Bi ai−1+l ∈ L(b; δ2j−2l−1 + e) Q l+1 2π3l m=1 ( k2 − j + m) i≥1

for l = 0, 1, . . . , j − 1 and all other coordinates are zero. Note, Ck+1 equals zero.

Odd Entry Case. This is a bit more technical. For n ≥ 2, consider the vector (0, . . . , Bn an , . . . , 0) where an is in the (2j + 1)-entry with all others zero, and Bn an ∈ L(b; δ2j+1 ). First, suppose n ≤ k/2 − j. Then 2j+1 entry

where

2(n−1+j)+1 entry z }| { z}|{ n ηj,n , . . . , 0) (0, . . . , Bn a , . . . , 0) = (0, 0, . . . , 0, C2(j+1) , 0, C2(j+2) , · · · )Gk + (0, . . . ,

ηj,n :=

n−2 Y l=0

3[k − (2(j + l) + 1)] Bn a ∈ L(b; δ2(n−1+j)+1 ) [2(j + l) + 1] 29

and for i = 1, 2, . . . , n − 1, we have C2(j+i) Next, suppose n ≥

k 2

−3i = π

Qi−2

l=0 [k − (2(j + l) + 1)] Qi−1 l=0 [2(j + l) + 1]

!

Bn an−1−i ∈ L(b; δ2(j+i) + e)

− j + 1. We will demonstrate that 2j+1 entry

z }| { (0, . . . , Bn an , . . . , 0) = (0, C2 , 0, C4 , 0, . . . , Ck , 0)Gk + kh

(6.2)

where h is given by (6.5) below and, for l = 1, 2, . . . , k/2, C2l =

Z 2k (k/2)!π

Bn an+j−l−1 ∈ L(b; δ2j + e)

where “Z” indicates some determinable integer. In (6.2), since the even numbered entries are all zero, let us delete them from this equation to obtain j th entry

z }| { ek + kodd h (0, . . . , Bn an , . . . , 0) = (C2 , C4 , . . . , Ck )G

(6.3)

where the vector on the left has (k/2) + 1 entries and Bn an is in the j-th entry, kodd is the vector k with the even ek is the (k/2) × (k/2 + 1)-matrix entries deleted, and G   2 − πa3 (k − 1)πa 2   (k − 3)πa − 3πa   3 e  . Gk =  .. ..  . .   2

− (k−1)πa 3

πa

Thus, we may rewrite this as

j+1 entry

z }| { (0, . . . , Bn an , . . . , 0) = (h, C2 , C4 , . . . , Ck )Nk Note, Nk is a square matrix. Furthermore, detNk = Lemma 6.13. detNk =

2k (k/2)! k/2 k π a 3k/2

where Nk :=



kodd ek G



.

by the following lemma.

k

2 (k/2)! k/2 k π a 3k/2

Proof. Expanding the determinant always using the left column, we get detNk =

 2  −1 k/2 π k/2 ak X k! k/2 k . 2j 3k/2 j=0 2k/2 · (k/2)! j

Thus, the lemma follows from the following general formula: −1 n  2  X n 2n (2n · n!)2 . = (2n)! j 2j j=0

(6.4)

To prove this last formula (of which I thank Zhi-Wei Sun for the following), notice that the above may be rewritten as  −1 −1 n  2  X n 2n n 2n =4 . j 2j n j=0

Therefore, since

we have (6.4) equal to

  2  −1    2n n 2n 2j 2n − 2j = n j 2j j n−j  n   X 2j 2n − 2j j=0

j

n−j

which is a well-known combinatorial formula.

30

= 4n

By Cramer’s rule, with Nk (1) denoting the matrix Nk with the first row replaced by (0, . . . , Bn an , . . . , 0), where Bn an is in the (j + 1)-entry, we have  !  k −1 k j 2 Y Y k 3 2 −j detNk (1) (6.5) (2i − 1)  = k [k − (2i + 1)] Bn an− 2 +j . h= detNk 2 (k/2)! i=1 i=j

Notice that h is divisible by a by our hypothesis that n ≥ k2 − j + 1. Using this and (6.3), it follows that kh ∈ Iker (b; 0) and Z C2l = k an+j−l−1 for l = 0, 1, 2, . . . , k/2. 2 (k/2)!π This finishes the proof of Lemma 6.12. Lemma 6.14. We have   (k) 1. Vk ∩ Ma (b)Gk = {0}

2. Iker (b) ∩ Vk = {0}   fa(k) (b)Gk = {0} 3. Iker (b) ∩ M

  (k) Proof. To prove the first statement, consider η ∈ Vk ∩ Ma (b)Gk . Then η1 = 2

−πa2 3 C2 .

Thus, η1 is divisible

2

by a which means η1 = C2 = 0. Next, η3 = (k − 1)πaC2 − πa C4 . Thus, C4 = 0. Continuing, we see that C2j = 0 for each j. Consequently, all the odd coordinates of η are zero. Next, since all the even coordinates of η (k) are constants and the image of Ma (b) by Gk is divisible by a, η must be zero. The second statement is clear. Lastly, let us prove the third. Suppose there exists h ∈ L(b) such that (C1 , C2 , . . . , Ck , 0)Gk = hk. Looking 1 at the last entry in this, we must have Ck = πa (Q>0 )h where “Q>0 ” is some positive rational number. Working 1 (Q>0 )hal for l = 0, 1, . . . , k/2 − 1. However, backwards from this using entries k − 1, k − 3, . . . , 3, we see Ck−2l = πa 2 1 1 k/2 the first entry says −πa which means C2 = − πa (Q>0 )hak/2 . This contradicts the positivity of 3 C2 = h 3k/2 a the rational constant. (k) f(k) Corollary 6.15. Ma (b) = Vk ⊕ M a (b)Gk ⊕ Iker (b). (k)

(k)

(k)

fa (b) and uGk ∈ Ma (b; ρ), then u ∈ M fa (b; ρ + e). Lemma 6.16. If u ∈ M 2

Proof. We will proceed by induction. Let ξ = uGk . Then ξ1 = − πa3 u2 ∈ L(b; δ1 ). Thus, u2 ∈ L(b; δ2 + e). Next, suppose u2j ∈ L(b; δ2j + e). Then there are p-adic units c1 and c2 such that ξ2j+1 = c1 πau2j + c2 πa2 u2j+2 ∈ L(b; δ2j+1 ). Since c1 πau2j ∈ L(b; δ2j+1 ), u2j+2 ∈ L(b; δ2j+2 + e) as desired. This finishes the odd coordinates of ξ. We now move on to the even. By the form of Gk , we can find p-adic units c1 and c2 such that ξk = c1 πauk−1 + c2 πa2 uk+1 ∈ L(b; δk ). f(k) However, since u ∈ M a (b), we have uk+1 = 0. Hence, uk−1 ∈ L(b; δk−1 + e). Next, suppose u2j+1 ∈ L(b; δ2j+1 + e). Again, we may find p-adic units c1 and c2 such that ξ2j = c1 πau2j−1 + c2 πa2 u2j+1 ∈ L(b; δ2j ).

It follows that u2j−1 ∈ L(b; δ2j−1 + e). Proceeding as in the previous section, Theorem 6.11 follows.

31

7

Dual Cohomology of Symmetric Powers

In this section, we will define a dual space Hk1∗ to the cohomology space Hk1 and a dual operator β¯k∗ to the Dwork operator β¯k . Dagger Spaces. Define b := analytic functions on 1 < |a| < e, e unspecified R

L† := analytic functions on |a| < e, where e > 1 unspecified

L†∗ := analytic functions on 1 < |a| ≤ ∞.

b×R b → Cp by hu, vi := the constant term of the product uv. By [14], this places Define a perfect pairing h·, ·i : R b into (topological) duality with itself. Further, by the Mittag-Leffler theorem, R b = aL† ⊕ L†∗ . Thus, since the R † b is aL† , the pairing places L† into duality with R/aL b annihilator of L† in R = L†∗ . ′ With any b , b > 0, define Ma := Ma (b′ , b) ⊗L(b′ ) L† † b M∗ := Ma (b′ , b) ⊗L(b′ ) R/aL . a

2

Fixing {x, x } as a basis of both Ma and M∗a , we may place them into (topological) duality by defining the pairing h·, ·, i : Ma × M∗a → Cp by taking h = h1 (a)x + h2 (a)x2 ∈ Ma and g = g1 (a)x + g2 (a)x2 ∈ M∗a and defining hh, gi := hh1 , g1 i + hh2 , g2 i. (k)

We may extend this duality, and the pairing, to the symmetric powers of these spaces as follows. Let Ma (k) (k)∗ (resp. Ma ) be the k-th symmetric tensor power of Ma (resp. M∗a ) over L† (resp. L†∗ ). For both Ma (k)∗ and Ma , fix the basis B := {v j wk−j }kj=0 where v := x and w := x2 . With (u1 , . . . , uk ) ∈ (Ma )k and (k)

(v1 , . . . , vk ) ∈ (M∗a )k , define a perfect pairing between Ma (k)∗ and v1 · · · vk ∈ Ma and defining (u1 · · · uk , v1 · · · vk ) :=

(k)∗

and Ma

(k)

by taking their images u1 · · · uk ∈ Ma

k X Y hui , vσ(i) i

σ∈Sk i=1

where Sk denotes the full symmetric group on {1, 2, . . . , k}. Differential Operator. With respect to the basis {x, x2 } of Ma , we may write the matrix representation of ∂a d as a da − G with   0 πa 2 . G := −πa 0 3 Note, we are thinking of Ma as a row space, and so, G acts on the right. Define the endomorphism of M∗a ∂a∗ := −a

d − T runca ◦ G. da

The map ∂a∗ acts on the column space M∗a . Similar to that of Da and Da∗ , ∂a and ∂a∗ are dual to one another with respect to the pairing h·, ·i. (k)∗ As we did in §5, we extend ∂a∗ to an operator on Ma by extending linearly the action ∂a∗ (u1 · · · uk ) :=

k X i=1

u1 · · · u ˆi · · · uk ∂a∗ (ui )

where u ˆi means we are leaving it out of the product. It is not hard to verify that ∂a∗ and ∂a are dual to one another via the pairing (·, ·). We may now define (k) Hk1 := M(k) and Hk1∗ := ker(∂a∗ |Ma(k)∗ ). a /∂a Ma (k)

(k)

While we have already defined Hk1 as the quotient space of Ma (b′ , b) by ∂a Ma (b′ , b), the two definitions are (k) (k) the same. Since Hk1 is of finite dimension, from [14, Proposition 7.2.2], ∂a Ma is a closed subspace of Ma . It 1∗ 1 follows that Hk is dual to Hk , and hence, they have the same dimension. 32

Dwork Operators. Similar to that of §5, define the Dwork operator (k) βk := ψa ◦ α ¯ k (a) : M(k) a → Ma .

Observe βk ◦ ∂a = p∂a ◦ βk , and so βk induces a map β¯k : Hk1 → Hk1 . With respect to the ordered basis B := (v k , v k−1 w, . . . , wk ), we may denote by Symk (A(a)) the matrix of α ¯ k (a). This makes the Dwork operator representable as βk = ψa ◦ Symk (A(a)). From this perspective, it is natural to define the dual of βk as βk∗ := T runca ◦ Symk (A(a)) ◦ Φa : Ma(k)∗ → Ma(k)∗ . Notice that since the matrix entries of A(a) lie in L† , βk∗ is well-defined. Dual to p∂k ◦ βk = βk ◦ ∂k is pβk∗ ◦ ∂k∗ = ∂k∗ ◦ βk∗ . Hence, βk∗ induces a map β¯k∗ : Hk1∗ → Hk1∗ . It follows that det(1 − β¯k T |Hk1 ) = det(1 − β¯k∗ T |Hk1∗ ).

7.1

Primitive Cohomology and its Dual

In this section, we will decompose the (conjugate) dual cohomology into three subspaces: a constant subspace, a trivial subspace, and a primitive part. The eigenvectors and associated eigenvalues of the dual Frobenius on the first two spaces are explicitly given. As we did in §2.5, we need to distinguish between π and its conjugate −π. Using notation from §2.5, for any b′ , b > 0, define the spaces

(k)

M−π,a := M−π,a (b′ , b) ⊗L(b′ ) L† † b M∗−π,a := M−π,a (b′ , b) ⊗L(b′ ) R/aL .

(k)∗

Denote by M−π,a and M−π,a the k-th symmetric tensor powers of these spaces over L† and L†∗ , respectively. (k)∗

(k)

d ∗ + Gk , and the differential operator on M−π,a is ∂−π,a := The differential operator on M−π,a is ∂−π,a := a da (k)

(k)

(k)∗

d 1 1∗ ∗ −a da + T runca ◦ Gk . From this, we may define H−π,k := M−π,a /∂−π,a M−π,a and H−π,k := ker(∂−π,a |M−π,a ).

Constant Subspace Ck+1 . Using the basis B defined above, define Ck+1 := p p k+1 1∗ Cp ⊂ H−π,k .

Lk

j=0

Cp v j wk−j . Notice that

Subspace Tk coming from infinity. (cf. §4.1) If k is odd, define Tk := 0. Assume k is even. Define κ := where i is a square root of −1 in Cp , and define
17
∞ 7 X 6 n 12 n v1 (a) := a−3n , n κn π n (n + 1)! 2 n=0

2i √ 3 3

and v2 (a) by replacing i with −i in v1 . By reduction, define the vectors in M∗a   1 −2 2 −1 ′ u1 := v1 v + − a v1 + 3a κπv1 + a v1 w 2   1 −2 2 −1 ′ u2 := v2 v + − a v2 − 3a κπv2 + a v2 w. 2 Next, for each 0 ≤ j ≤ k/2 and p|(k − 2j), define   3 3 ηj+ := a−k/2 e(k−2j)κπa u1k−j uj2 + e−(k−2j)κπa uj1 u2k−j   3 3 ηj− := a−k/2 e(k−2j)κπa u1k−j uj2 − e−(k−2j)κπa uj1 u2k−j .

If 4|k, then for each 0 ≤ j ≤ k/2 and p|(k − 2j), define ωj (a) by ωj (a2 ) = ηj+ (a). Else, if 2|k but 4 ∤ k, then for each 0 ≤ j < k/2 and p|(k − 2j), define ωj (a2 ) = ηj− (a). 1∗ of dimension It follows from §4.1 that Tk := spanCp {T runca (ωj (a))} is a subspace of H−π,k j k  k  1 + if 4|k  j k 2p k dimCp Tk = if 2|k but 4 ∤ k 2p    0 if k odd. 33

d b By Mittag-Leffler, T runca (ωj (a)) −Gk whose coordinates live in R. To see this, from §4.1 ωj (a) is a solution of a da
d is analytic on D− (∞, 1), and while a da − Gk T runca (ωj (a)) is not zero, it only consists of positive powers of ∗ the variable a. Hence, it is zero when we apply ∂−π,a .

Primitive Cohomology. Define the space 1∗ 1∗ P H−π,k := H−π,k /(Ck+1 ⊕ Tk ). p

(7.1)

1 1 1 Using this, we define the subspace P H−π,k ⊂ H−π,k as the annihilator of Ck+1 ⊕ Tk in H−π,k via the pairing. p 1∗ 1 Thus, P H−π,k and H−π,k are dual.

Frobenius on the Constant Subspace. Let us show ∗ β¯−π,k |Ck+1 = Symk (A(0)). p

(7.2)

∗ To prove this, recall β¯−π,k := T runca ◦ Symk A−π (a) ◦ Φa . The operator Φa is the identity map on Ck+1 and p k † Sym A−π (a) is a matrix with entries in L ; in particular, they are power series in the variable a. Therefore, only the constant terms of these power series will have an effect on Ck+1 , everything else would be killed by the p truncation operator. That is,

) = T runca ◦ Symk A−π (a)(Ck+1 ) T runca ◦ Symk A−π (a) ◦ Φa (Ck+1 p p ) = T runca ◦ Symk A(0)(Ck+1 p

∗ which proves (7.2). Further, since A(0) is invertible, we have β¯−π,k (Ck+1 ) = Ck+1 . p p

Frobenius on the Trivial Subspace Tk . Recall from §4.1 that A∗−π (a2 )a−p/2 V (ap )S(ap ) = a−1/2 V (a)S(a)M ! 3 eκπa 0 , and M is a constant matrix given by where V (a) is the matrix (u1 u2 ), S(a) := 3 0 e−κπa  √ p M= 0

(of course,

0 √ ± p



√ p may not be the positive square root) and   √ −g 0 √ M= ± −g 0

if p ≡ 1, 7 mod(12)

if p ≡ 5, 11 mod(12)

where “±” means positive if p ≡ 1, 5 mod(12) and negative if p ≡ 7, 11 mod(12), and g := g2 ((p2 − 1)/3). From this, we may calculate that for 4|k  k/2 p T runca (ωj ) if p ≡ 1mod(12)    (−1)j pk/2 T runc (ω ) if p ≡ 7mod(12) a j ∗ β¯−π,k (T runca (ωj )) = k/2  g T runc (ω ) if p ≡ 5 mod(12) a j    j k/2 (−1) g T runca (ωj ) if p ≡ 11 mod(12). and for 2|k but 4 ∤ k,

 k/2  p T runca (ωj )  (−1)j pk/2 T runc (ω ) a j ∗ β¯−π,k (T runca (ωj )) = (−g)k/2 T runca (ωj )    (−1)j+1 (−g)k/2 T runca (ωj )

if if if if

p ≡ 1mod(12) p ≡ 7mod(12) p ≡ 5 mod(12) p ≡ 11 mod(12).

∗ This demonstrates that not only is β¯−π,k stable on Tk , but the basis {T runca (ωj )} consists of eigenvectors. Using either Gauss sums or Dwork theory, one may show

∗ )= P−π,k (T ) := det(1 − β¯−π,k T |Ck+1 p

34

k Y

(1 − π1 (0)j π2 (0)k−j T )

j=0

where L(x3 /Fp , T ) = (1−π1 (0)T )(1−π2 (0)T ); note, we are using the (conjugate) splitting function exp(−π(t−tp )) ∗ to compute this L-function. Letting N−π,k (T ) := det(1 − β¯−π,k T |Tk ), we see that ∗

M k (T ) =

∗ 1∗ 1 det(1 − β¯−π,k T |H−π,k ) P−π,k (T )N−π,k (T )det(1 − β¯−π,k T |P H−π,k ) = 0 0 ¯ ¯ det(1 − pβ−π,k T |H ) det(1 − pβ−π,k T |H ) −π,k

−π,k

where we have used duality for the second equality and the bar on the left means complex conjugation. Note, conjugation acts as the identity on Mk∗ (T ) since it has real coefficients. It follows from (5.1) that Mk (T ) =

1 Nπ,k (T )det(1 − β¯π,k T |P Hπ,k ) 0 ¯ det(1 − pβπ,k T |H ) π,k

7.2

Functional Equation

1∗ 1 In this section, we will define an isomorphism Θk : P H−π,k → P Hπ,k which relates the Frobenius β¯π,k with its ∗ conjugate dual z¯−π,k ; see equation (7.7). (k)

For convenience, denote by A−π,k,a the matrix Symk (A−π (a)). Dual to β−π,k := ψa ◦ A−π,k,a : M−π,a →

(k)

(k)∗

(k)∗

(k)∗

∗ ∗ M−π,a is the isomorphism z−π,k := T runca ◦ A−π,k,a ◦ Φa : M−π,a → M−π,a . For ξ ∗ ∈ ker(∂−π,a |M−π,a ), we ±1 k+1 have (in Cp [[a ]] ) ∗ A−π,k,a ◦ Φa (ξ ∗ ) = z−π,k (ξ ∗ ) + η (7.3)

for some η ∈ Cp [a]k+1 . Recall the space R′−π,a (b′ , b) from §2.5. Define R′−π,a := R′−π,a (b′ , b) ⊗L(b′ ) L† . Define Θ′k,a := −a

d (k)∗ ∗ + Gk : ker(∂−π,a |M−π,a ) → Symk (R′−π,a ). da

∗ without the truncation map (and hence it p-commutes with It will be useful to note that Θ′k,a is both ∂−π,a ′ ′ A−π,k,a ◦ Φa ) and −∂π,a . Applying Θk,a to (7.3), we have (in Symk (R−π,a )) ∗ p A−π,k,a ◦ Φa ◦ Θ′k,a (ξ ∗ ) = Θ′k,a ◦ z−π,k (ξ ∗ ) + Θ′k,a (η).

(7.4)

¯ π (a) ◦ Recall from §2.5 the isomorphism Θ−π,a : R′−π,a (b′ , b∗ ) → Mπ,a (b′ , b) which satisfies Θ−π,ap = p−1 α Θ−π,a ◦ α ¯ ∗−π (a). Denoting by Θ−π,k,a the k-th symmetric power of Θ−π,a , and using that the matrix of α ¯∗−π (a) equals A−π (a), we have (7.5) pA−1 π,k,a ◦ Θ−π,k,ap = Θ−π,k,a ◦ A−π,k,a Thus, applying Θ−π,k,a to both sides of (7.4), we have ′ ∗ ′ ∗ ∗ ′ pk+1 A−1 π,k,a ◦ Θ−π,k,ap ◦ Φa ◦ Θk,a (ξ ) = Θ−π,k,a Θk,a z−π,k (ξ ) + Θ−π,k,a Θk,a (η)

which equals

′ ∗ ′ ∗ ∗ ′ pk+1 A−1 π,k,a ◦ Φa ◦ Θ−π,k,a Θk,a (ξ ) = Θ−π,k,a Θk,a z−π,k (ξ ) + Θ−π,k,a Θk,a (η).

Thus, ∗ Θ−π,k,a Θ′k,a (ξ ∗ ) = p−(k+1) βπ,k Θ−π,k,a Θ′k,a z−π,k (ξ ∗ ) + βπ,k Θ−π,k,a Θ′k,a (η).

Now, βπ,k Θ−π,k,a Θ′k,a (η) = ψa ◦ pΘ−π,k,ap (¯ α∗−π,k,a )−1 Θ′k,a (η)

α∗−π,k,a )−1 (η) = pΘ−π,k,a ◦ ψa ◦ Θ′k,ap (¯

α∗−π,k,a )−1 (η) = p2 Θ′k,a Θ−π,k,a ψa ◦ (¯

≡ 0 mod(∂π,a M(k) π,a )

(k)∗

∗ 1 where the first equality comes from (7.5). Finally, define Θk : ker(∂−π,k |M−π,a ) → Hπ,k by

Θk (ξ ∗ ) := Θ−π,k,a Θ′k,a (ξ ∗ ) Then (7.6) shows us that

mod(∂π,a M(k) π,a ).

−1 ∗ Θk = Θk z¯−π,k . pk+1 β¯π,k

However, this is not quite the functional equation since the map Θk has a kernel. 35

(7.6)

Lemma 7.1. ker Θk = Ck+1 ⊕ Tk . p ∗ ∗ Proof. Since z−π,k (Ck+1 ) = 0, we see by (7.6) that Θk (Ck+1 ) = 0. Next, since Θ′k,a is simply ∂−π,k without the p p ′ truncation map, we have Θk,a (Tk ) = 0, and so, Θk (Tk ) = 0. (k)∗

(k)

Suppose ξ ∗ ∈ M−π,a is in the kernel of Θk and is not constant. Then there exists h ∈ Mπ,a such that Θk (ξ ∗ ) = ∂π,a (h). Now, Θ−π,k,a Θ′k,a = Θ′k,a Θ−π,k,a , and so   d ˜ =0 − Gk (−ξ ∗ + h) a da ˜ := Θ−1 (h). Thus, −ξ ∗ + h ˜ corresponds to a solution in ker(lk |R). b But these solutions are in one-to-one where h −π,k,a correspondence with Tk . ∗ 1∗ 1 ∗1 1 Thus, if we restrict to z¯−π,k on P H−π,k and β¯π,k on P Hπ,k , then Θk : P H−π,k → P Hπ,k is an isomorphism which satisfies −1 ∗ Θk = Θk z¯−π,k . (7.7) pk+1 β¯π,k

1∗ ≤ k. Corollary 7.2. dimCp P Hk1 = dimCp P H−π,k 1∗ Proof. From the explicit description of the matrix Gk , the image of the injective function Θ′k : P H−π,k → Cp [a]k+1 k is contained within the subspace {0} × aCp which has dimension k.

8

Newton Polygon of Mk (T )

For k odd and k < p, from Theorem 6.3 and Section 7.1, we have Mk (T ) = det(1 − β¯k T |P Hk1 ) where

(k−1)/2

P Hk1 =

M

(Cp a)v k−2j w2j .

j=0

Using this explicit description of the cohomology, and the effective decomposition theorem (Theorem 6.3), we may prove the following. Theorem 8.1. Suppose k is odd and k < p. Writing Mk (T ) = 1 + c1 T + c2 T 2 + · · · + c(k+1)/2 T (k+1)/2 we have ord(cm ) ≥

(p − 1)2 2 (m + m + mk). 3p2

for every m = 0, 1, . . . , (k + 1)/2. Proof. Let b := (p − 1)/p and b′ := b/p. From §2.6, we know α(a)xi ≡ Ai1 (a)x + Ai2 (a)x2 mod(Da K(b′ , b)) with ′ (k−1)/2 Aij ∈ L(b′ ; b3 (pj − i)). From §6.2, we know {av k−2i w2i }i=0 is a basis of P Hk1 . Since v := x and w := x2 , α ¯ k (a)(av k−2i w2i ) = a(¯ α(a)v)k−2i (¯ α(a)w)2i =

k X

aBr (a)v k−r wr

r=0

where Br (a) :=

X

n+m=k−r n=0,1,...,k−2i m=0,1,...,2i

   k − 2i 2i (A11 )n (A21 )k−2i−n (A12 )m (A22 )2i−m . n m

After some calculation, we see that aB2r ∈ L(b′ ;

b′ b 2b′ b′ [(p − 1)k + pr − 2i] − 2b′ /3) = L(b′ ; [(p − 1)k − 2i] + δr+1 − − ). 3 3 3 3

36

where δr+1 :=

b(r+1) . 3

Using notation from §6.2, this means (k)

α ¯ k (a)(av k−2i w2i ) ∈ Map (b′ ; δ + Consequently,

b 2b′ b′ [(p − 1)k − 2i] − − ). 3 3 3

b′ b 2b′ [(p − 1)k − 2i] − − ). 3 3 3 we may write

βk (av k−2i w2i ) ∈ M(k) a (b; δ + It follows from Lemma 6.7 that for some constants Aij (k−1)/2

X

βk (av k−2i w2i ) ⊂ with the sum being an element of Vk (b; δ +

b′ 3 [(p

j=0

Aij av k−2j w2j + ∂a M(k) a (b)

− 1)k − 2i] −

Aij a ∈ L(b; δ2j+1 +

b 3

−

2b′ 3 ).

Therefore,

b 2b′ b′ [(p − 1)k − 2i] − − ), 3 3 3

and so,

2b′ b′ 2 (pj − i) + (p − 1)k + (b − b′ ). 3 3 3 ′ ˜ Let us rewrite this. Let ξ ∈ Cp such that ordp (ξ) = 2b /3. Fix ξ := ξ p−1 . Then, if we had used the basis (k−1)/2 , then {ξ i av k−2i w2i }i=0 (k−1)/2 X
ξ i−j Aij ξ j av k−2j w2j . β¯k (ξ i av k−2i w2i ) = ord(Aij ) ≥

j=0

Hence, the matrix of β¯k on P Hk1 , with respect to this new basis, takes the form 

 ′  b′ 2 matrix of β¯k = p 3 (p−1)k+ 3 (b−b )  

B00 B10 .. .

˜ 01 ξB ˜ 11 ξB .. .

··· ··· .. .

B(k−1)/2,0

˜ (k−1)/2,1 ξB

···

ξ˜(k−1)/2 B0,(k−1)/2 ξ˜(k−1)/2 B1,(k−1)/2 .. . ξ˜(k−1)/2 B(k−1)/2,(k−1)/2

where ξ˜j Bij = ξ i−j Aij , and so, ordp (Bij ) ≥ 0. It follows that, if we write



  , 

(8.1)

det(1 − β¯k T |P Hk1 ) = 1 + c1 T + · · · + c(k+1)/2 T (k+1)/2 then since b := (p − 1)/p and b′ := b/p, ord(cm ) ≥

(p − 1)2 2 (m + m + mk). 3p2

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C. Douglas Haessig Department of Mathematics University of Rochester Rochester, NY 14627

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